Chapter one Introduction Energy sources The various forms

NOTES ON ELECTRIC CIRCUIT AND MACHINES Prepared By Masen

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Chapter one Introduction

Energy sources

The various forms of energy available in nature are as follow.

Radient Energy:Energy obtained from sunlight

Thermal Energy: Energy obtained from earth’s interior

Chemical Energy: Energy store in wood,coal & oil in chemical form

Potential Energy: Energy obtained by virtue of position.eg:bulk of water stored in

high level pressure energy

Kinetic Energy: Energy obtained from moving bodies

Nuclear Energy: Energy release when atoms are broken down

Electrical energy

Production or generation of energy

The practical device used for the generation of electrical energy is generator.

Generator is rotating device which converts the mechanical energy into electrical energy.

In this process mechanical power is supplied to shaft in form of torque. The mechanical

torque produced by turbine is converted into electrical energy. The turbine used may be of

two type:Hydraulic and steam driven turbine. The electrical source may be thermal and

non thermal.

The various types of thermal and non thermal power generation schemes are as

follows:

Thermal power

Coal feed thermal power generation

Diesel power generation

Nuclear power generation

Solar power generation

Non-thermal power

Hydropower generation

Tidal power generation

Wind power generation

Coal feed thermal power generation

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It is the type of generator that uses coal for the generation of steam from the boiler.

This steam is then passed into steam turbine through super heater. The steam turbine

converts the energy of steam into mechanical rotation. The shaft of turbine is connected to

the generator which generates the electrical energy. The discharge from the steam turbine

is passed through a condenser in which steam is passed through a condenser. Here steam is

condensed into water and this water ia again supplied to boiler with help of field pump. In

this type of power generation the starting and warming up process is comparetively long

therefore frequently shutting up and starting up is not suggestable.

Diesel power generation

This type of power station uses diesel as fuel in diesel engine. The mechanical

rotation produced by diesel engine is coupled with generator by the help of a shaft. This

generator then converts the mechanical force into electrical energy. The engine has to be

cooled against the heat generation within the engine so, circulating water system is used

for cooling purpose.

Nuclear power generation

The nuclear fission of atomic material such as Uranium produces heat which is

utilized to produce steam to run steam turbine.Such power station is called nuclear power

generation. About 3000 netrictones of coal produce the same amount of heat as 1kg of

nuclear fuel when uranium 225 is bombarded with nutrons. Fission reaction takes place

releasing nutrons and heat energy. These neutrons then participate in chain reaction of






emissioning atoms of 285 uranium. The nuclear reactor of this plant has to be carefully

shield against the radiation.

Solar power generation

In this type of power generation, solar energy is collected directly with the help of

solar panel which converts it into electrical energy. Solar panel consists of solar cell which

are semicinductor devices. Number of solar cells can be connected in series parallel

combination to generate electrical power at desired level. The energy generated by solar

pannel is used to charge battery which gives DC voltage output. Inverter may be used to

convert the voltage into ac if the user needs ac voltage.

Hydropower Generation

Hydropower is economical and pollution free source of energy. The initial capital

investment in dams, transmission and generation is quite high but operating cost is very

low. A tyical layout of storage type hydropower is given below

Hydropower scheme needs a supply of water and difference in water level so that

the water will possess potential energy. If certain amount of water can be collected at

certain height above a certain reference level. Then the potential energy possessed by this

amount of water can be converted into mechanical rotation and thus into electrical energy.

The power generated by a hydropower plant is given by,

P=Q x g x ρ x H =9.81x1000xQxH

∴ P = 9810QH

Where,

Q=discharge of water (in m3/sec)

H=height of water that falls through certain height (in meter)

The actual power is always less than this theoretical value because of various

power loss such as friction losses in pipe, losses in turbine, losses in generation etc.






Tidal power energy

This type of power plant is used to convert the change in potential energy in

potential energy caused by tide levels into electrical form. A dam with large gates is made

across the mouth of bay in ocean and low head turbines are used. At the time of high tides,

gates are opened and after water reaches to electrical level the gates are closed. The water

so collected is used to drive the turbines that in turn rotates the generators converts it into

electrical energy.

Wind power energy

In this type of generation, wind is used to drive the air turbines which in turn

rotates the generator and the generator then converts the mechanical rotation into

electricity according to faradays laws of electromagnetic induction.

Generation, distribution and transmission of electric process

Above figure shows the layout of power distribution system consisting of generation

transmission and distribution section. The power system consist of two power station

system. Most of the power station are quite far from the city so the power generation in

power station has to be transmitted through a transmission line. In order to maintain lower

power loss the transmission is done at high voltage level. Set up transformer are used to

step up the voltage level generated by the power station and then transmitted through

transmitted line. The step down transformer are used at substation to step down the voltage

before distribution.

The 11kV distribution lines run around the city and 400V 4 wire and 220V single

phase system are trapped out from the distribution line at various points by using

11kV/400V step down transformer.






Chapter two Passive elements and laws in electrical engineering

Electric current and emf

Electric current may be defined as time rate of net motion of an electric charge

across a cross sectional boundary.

Fig: 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦

Rate of transfer of electric energy =Quality of electric charge transfer

Time duration

=𝑑𝑄

𝑑𝑡 (c/sec) Ampere

Electromagnetic Force(emf) is the force that causes on electric current to flow in an

electric circuit while the potential between two point exist. Volt is unit of emf & P.d

(potential difference)

Resistance

It may be defined as the property of a substance due to which it opposes (or

restricts) the flow of electricity (i.e., electrons) through it. Commonly it is denoted by R.

The Unit resistance is ohm(Ω).

A conductor is said to have a resistance of

one ohm if it permits one ampere current to flow

through it when one volt is impressed across its

terminals.

Laws of Resistance

The resistance R offered by a conductor depends on the following factors :

(i) It varies directly as its length L. i.e. RαL ……………………………………..i

(ii) It varies inversely as the cross-section A of the conductor. i.e. Rα1

A……......ii

(iii) It depends on the nature of the material.

(iv) It also depends on the temperature of the conductor.

From Equation i & ii

RαL

A

∴R=ƍL

A ……………………………………………………………………..…iii

Where ƍ is a constant depending on the nature of the material of the conductor and is

known as its specific resistance or resistivity of conductor

If in Eq. (iii), we put L=1m and A =1m2, then R = ƍ

Hence, specific resistance or Resistivity of a material may be defined as the resistance

between the opposite faces of a meter cube (m3) of that material. Its unit is Ωm.

Conductance and Conductivity

Conductance (G)is reciprocal of resistance. Whereas resistance of a conductor measures

the opposition which it offers to the flow of current, the conductance measures the

inducement which it offers to its flow.

G= 1

𝑅 =

1

ƍ𝐿

𝐴

= 1

ƍ𝐴𝐿= σ 𝐴𝐿






Where σ is conductivity or specific conductance of a conductor. The unit of

conductance is Siemens (S). Earlier, this unit was called mho.

Resistance variation with Temperature

Inferred zero temperature shows that at -t˚c is the temperature at which resistance

would be zero. If the rate of decrease between 100˚c & 0˚c were maintained constant at all

temperature. This doesn’t mean that resistance of the metal is actually zero at that

temperature.

From similarity of triangle

R2

R1=

t0+t2

t0+t1

Where R1 and R2 are resistance at which t1 and t2 respectively.

Temperature Coefficient of Resistance

Let a metallic conductor having a resistance of Ro at 0°C be heated of to 0°Cand let its

resistance

at this temperature be Rt, Then, Rt

R0=

t0+t

t0+0=1+

t

t0

∴Rt=R0(1+α0t)

Where α0=1

t0 is called temperature of coefficient of resistance of material at 0° C

And change in resistance is given by

∆R=Rt-R0

=R0(1+α0t)-R0

=R0α0t

Temperature Coefficient of Resistance is defined as the ratio of increase in resistance

per degree rise in temperature the original resistance.

It is found that the Temperature Coefficient of Resistance depends

(i) directly on its initial resistance

(ii) directly on the rise in temperature

(iii) on the nature of the material of the conductor.

Effect of Temperature on Resistance:

(i) For all pure metals temperature of coefficient is positive i.e. increase the temperature

increases resistance of pure metals.






(ii) Increase in temperature increase the resistance of alloys, though in their case, the

increase is relatively small and irregular. For some high-resistance alloys like Eureka

(60% Cu and40% Ni) and manganine the increase in resistance is (or can be made)

negligible over a considerable range of temperature.

(iii)Increase in temperature decrease the resistance of electrolytes, insulators (such as

paper, rubber, glass, mica etc.) and partial conductors such as carbon. Hence, insulators

are said to possess a negative temperature-coefficient of resistance.

Ohm's Law , its application and limitation

Ohm’s law states that the ratio between potential difference applied across a

conductor and the current passing through it is always constant and is equal to value of

resistance ot that circuit element provided physical state remains constant.

i.e. V

I =constant=R

where R is known as resistance of conductor.

In figure the resistance R is supplied by

a voltage source. The voltmeter measures

voltage across the resistance and ammeter

measures current passingthrough the resistance.

When voltage across the resistance is increased,

the current through the resistance increases in

same proportion.

Application of ohm’s law

Basic tools for the electric circuit analysis

To calculate resistance of any circuit

To measure current through resistance

Limitation of ohm’s law

It doesn’t apply to the circuit whose temperature varies at different values of current

passing through it.

Resistance in Series

When some conductors having resistances

R1,R2and R 3 etc. are joined end-on-end so that

same current passes through all of them then they

are said to be connected in series

conductors.

V = V1+ V2+ V3= IR1 + IR2+ IR3

But V = IReqv

where R is the equivalent resistance of the series

combination.

IReqv= IR1 + IR2+ IR3

or Reqv= R1 + R2+ R3

Also1

𝐺𝑒𝑞𝑣=

1

G1+

1

G2+

1

G3

As seen from above, the main characteristics of a series circuit are:

1. Same current flows through all parts of the circuit.






2. Voltage drop across different resistors have their individual voltage drops due to its

different resistance and is given by Ohm's Law

3. Voltage drops are additive.

4. Applied voltage is equals the sum of different voltage drops.

5. Resistances are additive.

6. Powers are additive.

Voltage Divider Rule

Since in a series circuit, same current flows through each of the given resistors, voltage

drop varies directly with its resistance. In above Fig.

Voltage drop across R1 is V1=IR1= V

ReqvxR1

Voltage drop across R2 is V2=IR2= 𝑉

Reqv𝑥R2

Voltage drop across R3 is V3=IR3= 𝑉

Reqv𝑥R3

Resistances in Parallel

When resistances,as joined in such way that current

divides into two or more paths from a node then the

resistors are said to be connected in parallel.

Here

I=I1+I2+I3=V

R1+

V

R2+

V

R3…………………..(i)

We have I= 𝑉

𝑅𝑒𝑞𝑣

Now equation (i) becomesV

Reqv=

V

R1+

V

R2+

V

R3

or1

Reqv=

1

R1+

1

R2+

1

R3

∴Reqv = R1R2R3

R1R2+R2R3+R1R3

The main characteristics of a parallel circuit are :

1. Same voltage acts across all parts of the circuit

2. Different resistors have their individual current.

3. Branch currents are additive.

4. Conductances are additive.

5. Powers are additive.

Current Divider Rule






For fig:a

I1= IR1

R1+R2

I2= IR2

R1+R2

Power and energy in dc circuit

Let us consider a dc circuit as shown in figure which

consists of resistance R and dc voltage supply V. in the

circuit I flows through load resistance Rl and voltage drives

some charge q through circuit against opposition offered by

resistance Rl

. Energy delivered to load=V.q…………………………(i)

From defination of electric current q= I.t………………(ii)

Now from equation (i)&(ii)

E=VIt…………………………………………………..(iii)

Power is defined as time rate of energy delevering capacity of source.

Power (p)= E

T =

VIt

𝑇 = VI

Since V=IR

P=I2R

Active and passive elements

Active elements are those elements which supply energy to the networks.

Examples: voltage sources like batteries, dc generators etc.

The elements which dissipates or store energy are known as passive elements.

Example: resistors, capacitors etc.

Inductance

Inductance is the property of the circuit elements which oppose the rise or fall of

the current through it by inducing emf across the circuit elements. The circuit having these

property is known as inductor. It is capable of storing finite amount of energy. Generally

a wire wound in form of coil has these property.

Let us consider an inductive coil carrying a time varying

current ‘i’ as shown in fig.

The emf induced in the coil is given by E=Ldi

dt

Where L is inductance of coil and can be expressed as L=edi

dt

∴i=1

L∫ edt

Inductance can be defined as capacity to induce emf per unit time rate of change of

current. A coil having larger value of inductance can induce a large value of emf for given

value of 𝑑𝑖

𝑑𝑡

M=k√𝐿1𝐿2 where k= coupling coefficient






Inductance in series

Fig:series adding connection Fig:series opposing connection

Let M = coefficient of mutual inductance

L1= coefficient of self-inductance of lst coil

L2= coefficient of self-inductance of 2nd coil

For series adding connection, the total emf induced in each of coils L1& l2 is due to coil

Self inductance & emf induced by other coil called mutual inductance

i.e. e1=L1di

dt+ M

di

dt

similarly

e2=L2di

dt+ M

di

dt

et= (L1 + L2 + 2M)di

dt

Total inductance =𝑒𝑡𝑑𝑖

𝑑𝑡

= L= L1 + L2 + 2M

For series opposing connection

Total inductance = L= L1 + L2 − 2M

Inductanc in parallel

When mutual flux helps the individual flux

L =L1L2 − M2

L1L2 − 2M

When mutual flux opposes the individual flux

𝐿 =L1L2 − M2

L1L2 + 2M

Capacitor

A capacitor is a circuit element which is capable of storing and delivering finite

amount of charges consisting of two conducting surfaces separated by a layer of an

insulating medium called dielectric. The conducting surfacesmay be in the form of either

circular (or rectangular)plates or be of spherical or cylindrical shape. The purpose of a

capacitor is to store electrical energy by electrostatic stress in the dielectric

Capacitance

The property of a capacitor to 'store electricity' may be called its capacitance.In

other word the capacitance of a capacitor is defined as "the ratio of magnitude of the total

charge Q on either side conductor to the potential difference V between the conductor.Its

unit is farad (F)

C=𝑄

𝑉

Q=CV

Differentiating both side with respect to t dq

dt =C

dv

dt i=C

dv

dt

∴C= i

dv

dt






The voltage across the capacitor of any instant is given by

V=1

c∫ idt

Capacitance in series

V= Q

C

V1= Q

c1 V2=

Q

c2 V3=

Q

c3 (1)

Now,

V=V1+V2+V3

𝑜𝑟,Q

C=

Q

c1 +

Q

c2 +

Q

c3

∴1

C=

1

c1+

1

c2+

1

c3

∴ C = C1C2C3

C1.C2+C1.C3+C2.C3

We can also find the avlue of V1,V2 & V3.from (1) we get

Q=V1C1=V2C2=V3C3 =VC

or,V1C1= VC=VC1C2C3

C1.C2+C1.C3+C2.C3

∴ V1= V𝐂𝟐𝐂𝟑

𝐂𝟏.𝐂𝟐+𝐂𝟏.𝐂𝟑+𝐂𝟐.𝐂

𝟑

∴ V2= V𝐂𝟏𝐂𝟑

𝐂𝟏.𝐂𝟐+𝐂𝟏.𝐂𝟑+𝐂𝟐.𝐂

𝟑

∴ V3= V𝐂𝟐𝐂𝟏

𝐂𝟏.𝐂𝟐+𝐂𝟏.𝐂𝟑+𝐂𝟐.𝐂

𝟑

Capacitance in parallel

V= 𝑄

𝐶 → Q = CV

If Ceqv be equivqlent capacitance then

Q=Q1+Q2+Q3

Or Q=C1V+C2V+C3V

Or 𝑄

𝑉 =C1 +C2+C3

∴ Ceqv=C1 +C2+C3

Power and torque relation

Let us consider a rotating system rotating under a couple of force ‘f’ as shown

below.

Let,

r =radius of rotating disk in rpm

N=speed of rotating disk in rpm

Time for one revolution=60

𝑁

Distance covered in 1 sec= 2πr

Now,

Torque developed(T)=fxr

We know that,

Work done= force x distance

∴ 𝑤 = 𝑓𝑥2πr

∴ power(p) =work done

time taken=

f x 2πr60

N

=Tx2πN

60

∴ 𝑃 =Tx2πN

60

This equation shows that for a constant power of the system, more torque can be

developed at lower speed.






Kirchoff’s Laws

These laws are more comprehensive than Ohm's law and are used for solving electrical

networks which may not be readily solved by the latter. Kirchoff’s laws, two in number,

are particularly useful

(a) In determining the equivalent resistance of a complicated network of conductors and

(b) For calculating the currents flowing in the various conductors. The two-laws are :

1. Kirchoff’s Point Law or Current Law (KCL)

It states as follows:

In any electrical network, the algebraic sum of the currents meeting at a point (or

junction) is zero.

In another way, it simply means that the total current

leaving a junction is equal to the total current entering

that junction.

Consider the case of a few conductors meeting at

a point A as in Fig. kcl 1 . Some conductors have

currents leading to point A, whereas some have currents

leading away from point A. Assuming the incoming

currents to be positive and the outgoing currents

negative, we have

I1+ (-I2) + (-I3) + (+ I4) + (-I5) =0

I1 - I2-I3+ I4-I5) =0

I1+ I4=I2+I3+ I5

or incoming currents =outgoing currents

Similarly, in Fig. KCL 2,

for node A

+I + (-I1) + (-I2) + (-I3) + (-I4)=0

or I=I1+ I2+ I3+ 14

We can express the above conclusion thus:

∑I= 0 at A junction

2.Kirchoff’s mess law or Voltage Law (KVL)

It states as follows :

The algebraic sum of the products of currents and resistances in each of the

conductors in any closed path (or mesh) in a network & the algebraic sum of the e.m.f’s.

in that path is zero.In other words,

∑IR + ∑e.m.f. = 0 ...round a mesh

It should be noted that algebraic sum is the sum which

takes into account the polarities of the voltage drops.

In Fig: KVL a closed circuit with resistor R1 ,

R2 & R3 are supplied with a voltage source V. I is

current in the circuit. The direction of the current is

always from negative to positive within the battery.

Sign convention

i. +ve to entering point of the resistance and –ve for the

leaving point






ii.Potential difference measured from the –ve to +ve is called rise in voltage or +ve

voltage drop. Potential difference measured from the +ve to -ve is called fall in voltage or

-ve voltage drop.

Applying kirchhoff’s law

+V+(-V1)+(-V2)+(-V3)=0

Or V-IR1-IR2-IR3=0

∴ V=IR1+IR2+IR3

Note:It is important to note that

1.The sign of the battery e.m.f is independent of the direction of the current through that

branch.

2.The sign of voltage drop across a resistor depends on the direction of current through

that resistor but is independent of the polarity of any other source of e.m.f in the circuit

under consideration.

Use of kirchoff’s law

Determining equivalent resistance of a complecated network of conductor

Calculating current flowing in various conductor

Application of kirchoff’s law using mesh analysis

To perform mesh law analysis the total number of mesh in current is identified

then current for each mesh is given. The direction of all mesh currents are taken as

closewise. The KVL is applied on each mesh and finally the current at various branches

can be calculated by solving the KVL equations obtained for each mesh.

Using KVL in mesh I

V1 − I1R1 − (I1 − I2)R4 = 0

∴ V1 − I1(R1 + R4) + I2R4 = 0 ……………..(i)

Using KVL in mesh II

−(I2 − I1)R4 − I2R2 − (I2 − I3)R5 = 0

∴ I1R4 − I2(R4 + R2 + R5) + I3R5 = 0 ………………(ii)

Using KVL in mesh III

−(I3 − I2)R5 − I3R3 − 𝑉2 = 0

∴ I3(R3 + R5) + I2R5 − 𝑉2 = 0 ……………….(iii)

Solving equation (i),(ii)&(iii) we can find the current at various branches and loop.






Application of kirchoff’s law using nodal analysis

In nodal analysis, the

number of nodes in an

electrical network is identified

then one of node is regarded

as reference node which has

zero potential. If there are n

nodes, the equation for (n-1)

nodes are obtained. These

nodal equation are then simplified in order to obtain various branches current.

Using KCL in node1

I1 = I2 + I4 V1−VA

R1=

VA−VB

R2+

VA

R4 ……(i)

Using KCL in node 2

I5 = I2 + I3 VB

R5=

VA−VB

R2+

V2−VB

R3 ……(i)

Solving equation (i)&(ii) we get the values of VA and VB and finally the value of currents

at the node can be obtained

Faradays law of electromagnetic induction

Frist Law:It states :

Whenever the magnetic flux linked with a circuit changes, an e.m.f. is always induced in

it.

or

Whenever a conductor cuts magnetic flux.. an e.m.f. is induced in that conductor.

Second Law. It states :

The magnitude of the induced e.m.f. is equal to the rate of change of flux-linkages.

Explanation. Suppose a coil has N turns andflux through it changes from an initial value

of Φ1webers to the final value of Φ2 webers in time t seconds. Then. remembering that by

flux-linkages is meant the product of' number of turns by the flux linked with the coil, we

have.

Initial flux linkages= NΦ1

Final nux linkages= NΦ2

∴induced emf (e) =NΦ2−NΦ1

t web/m or volt

Putting the above expression in its differential form, we get

E= d

dt(NΦ)

E =Nd

dtΦ volt

Induced e.m.f

Induced e.m.f can be either(i) dinamically induced or (ii) statically induced

1.Dynamically induced emf:

Let us consider a conductor AB placed in magnetic field as shown in the figure. When the

conductor is moved upward it will cut the magnetic flux.






Therefore according to Faradays law of electromagnetic induction, emf will induced in

conductor AB

Let

L=length of conductor

V=velocity of conudctor

B=magnetic flux density

Then

Distance moved in small time dt =dx=Vdt

Area cut by conductor in time dt=𝐿 𝑑𝑥𝐵

Rate of change of flux= L dxB

dt

Hence emf =L dxB

dt=

L VdtB

dt =BLV

If conductor moves making angle θ with direction of magnetic flux then emf

induced is given by

E=BLVsin θ

Statically induced

There is no related relation between conductor and magnet.Coil A is supplied by

Dc voltage source with a variable resistance in series & coil B is connected with

galvanometer. When current through coil A is varied by variable resistance the

galvanometer will show some deflection indicating induced emf in coil B. flux linkage

in coil will change with respect to time hence emf will induced in B according to

faraday’s law of electromagnetic induction

Let coil has N numbers of turns. Magnetic flux changes from Ф1 to Ф2 in t second.

Initial flux linkage =NФ1

Final flux linkage =NФ2

∴ induced emf = rate of change of flux linkage

=NФ2−NФ1

t

E=NdФ

dt

Direction of induced emf

The direction of induced emf is determined by lenz law. According to lenz law the

statically induct ef will drive the current in such a way direction that magnetic flux

produced by induct current will oppose the cause by which emf was induct.

SOLVED NUMERICAL PROBLEMS

Q.No.1. A platinum coil has a resistance of 3.14 Ω at 40°C and 3.767 Ω at 100°C.

find the resistance at 0°C and the temperature – coefficient of resistance at 40° C.

soln,given

R40=3.14 Ω R100=3.767 Ω R0=? α0=?

We have, R100

R40=

1+100α0

1+40α0

or,3.767

3.14=

1+100α0

1+40α0 α0=0.00379per °C

now, α40 =α0

1+40α0=

0.00379

1+40x0.00379

∴ α40 = 3.29x10−3






Q.No.2. What is the value of unknown

resistor R in fig if the voltage drop across the

500 resistor is 2.5 volts?All resistor are in Ω.

soln

By direct proportion method,

Voltage drop on 50 Ω =2.5 x50

500= 0.25 V

Voltage drop on CD=0.5+2.5=2.75 Ω

Voltage drop across 550 Ω = 12V – 2.75= 9.25 V

Current on 550 Ω resistor= V550

550=

9.25

550= 0.0168A

Current across 50 Ω resistor= 0.25

50= 0.005A

∴ Current across R=0.0168 – 0.005=0.0118A

Hence, R= V

I=

2.75

0.0118= 233 Ω

∴ R = 233 Ω

Q.No.3. Find the equivalent resistance between

point X & Y of the given circuit. Also find how

much current will pass through each

resistance?

soln

The given circuit can be re-arrange as follow

As 60 Ω ,40 Ω &12 Ω are in parallel

1

Reqv=

1

60+

1

40+

1

12=

40x12 + 60x12 + 50x60

60x40x12=

1

8

∴ Reqv=8 Ω

Total current through the circuit (I) = V

R=

24

8= 3A

I1 = 3x40x12

40x12+60x12+60x40=0.4A

I2 = 3x60x12

40x12+60x12+60x40=0.6A

I3 = 3x40x60

40x12+60x12+60x40=2A

Hence 0.4A,0.6A & 2A current will pass through 60 Ω ,40 Ω &12 Ω resistance

respectively.

Q.No.4. Find the value of total current and power in the circuit shown below.






Reqv = 1 + 2 + 4||4 + 3 + 5 = 3 +4𝑥4

4+4+ 8 = 11 + 2

∴ Reqv = 13 Ω

Now, V=IR

0r,65= 13 x I

∴ I =65

13= 5A

P=VI=65 x 65

13=325w

Q.No.5. Find the resistance between the point A & B shown below.

Soln,

The equivalent resistance between point EF is

𝑅𝐸𝐹 = (6 + 10 + 6)||10 = 22||10 =22x10

22+10= 6.875𝛺

Now equivalent circuit became as in fig:A

The equivalent resistance between CD is given by

𝑅𝑐𝑑 = (6 + 6.875 + 6)||15 = 18.875||15 +18.875x15

18.875+15= 8.357𝛺

The circuit becomes as in fig:B

The equivalent resistance between AB is

RAB = (10 + 8.357 + 6) Ω = 24.357Ω

Q.No.6. If 20V be applied across AB shown in fig. calculate the total current,the

power dissipited in each resistor and the value

of series resistance to have total current.

Soln

As a&b are in parallel their equivalent resistance is

Rab= 2 // 4= 2x4

2+4=

4

3Ω

Also

Rcd= 6 // 8= 6x8

6+8=

24

7Ω

Again Rab & Rcd are in parallel hence

Rabcd= 4

3∥

24

7 =

4

3x24

74

3+

247

= 2425

Ω






f & g are in parallel hence Rfg=3 ∥ 6 =3x6

3+6=

18

9= 2Ω

now,given circuit becomes as shown in fig beside

now,Reqv=(24

25+ 2) || 5 =

74

25 || 5 =

370

199 M

∴ Reqv= 370

199

Hence total current (Itotal)= V

Reqv=

20370

199

Itotal=10.76A

I1=𝑉

𝑅𝐴𝑀𝐵=

2074

25

=6.76A

I2=10.76 – 6.76=4A

Now voltage drop across Resistance a,b,c& d (Vabcd)=IRabcd=6.76X24

25= 6.48V

Ia=Vabcd

Ra=

6.48

2= 3.24A

Ib=Vabcd

Rb=

6.48

4= 1.62A

Ic=Vabcd

Rc=

6.48

6= 1.08A

Id=Vabcd

Rd=

6.48

8= 0.81A

Ie=V

Re=

20

5= 4A

If=I1𝑋𝑅𝑔

𝑅𝑓+𝑅𝑔= 6.76𝑋

6

3+6= 4.51A

Ig=I1𝑋𝑅𝑓

𝑅𝑓+𝑅𝑔= 6.76𝑋

3

3+6= 2.25A

Power dissipatation

Pa=𝐼𝑎2𝑥𝑅𝑎=3.242x2=20.99W

Pb=𝐼𝑏2𝑥𝑅𝑏=1.622x4=10.4W

Pc=𝐼𝑐2𝑥𝑅𝑐=1.082x6=7W

Pd=𝐼𝑑2𝑥𝑅𝑑=0.812x8=5.25W

Pe=𝐼𝑒2𝑥𝑅𝑒=42x5=80W

Pf=𝐼𝑓2𝑥𝑅𝑓=4.512x3=61W

Pg=𝐼𝑔2𝑥𝑅𝑔=2.252x6=30.4W

The value of series resistance to have total current is Reqv= 370

199 Ω

Q.No.7.Solve the following circuit using kirchoff’s laws(KVL or mesh analysis

method) and find current through each branch.

soln

Using KVL in ABED

+50 − 10I1 − 10I1 − (I1 − I2)10 −20 = 0

0r,30 − 30I1 + 10I2 = 0

∴ 3I1 − I2 = 3 …………………(i)

Using KVL in BCFE

20 − 10(I2 − I1) − 5I2 − 5I2 −40 = 0

0r,−2 − 2I2 + I1 = 0

∴ I1 − 2I2 = 2 ……………….(ii)

Solving equation (i)&(ii)

I1 = 0.8A

I2 = −0.6A

Hence current in resistor of 10Ω in branch AD & AB is given by

I1 = 0.8A

Current in resistor of 10Ω in branch BE is I1 − I2 = 0.8A + 0.6A = 1.4A

Currrent in resistor of 5Ω at branch BC & CF is given by I2 = −0.6A

i.e current flow from F-C & C-B






Q.No.8. Determine the current supplied by each battery in the circuit shown below

by applying KVL( mesh analysis method).

Soln

Using KVL in mesh I

20 − 5I1 − 3(I1 − I2) − 5 = 0

∴ 8I1 − 3I2 = 15 ………….(i)

Using KVL in mesh II

5 − 3(I2 − I1) − 4I2 + 5 − 2(I2 − I3) + 5 = 0

∴ 9I2 − 3I1 − 2I3 = 15 ………… (ii)

Using KVL in mesh III

−5 − 2(I3 − I2) − 8I3 − 30 = 0

∴ 2I2 − 10I3 = 35 ……..(iii)

Solving equation (i),(ii)&(iii) we get

I1 = 2.56A

I2 = 1.82A

I3 = −3.13A Now,

Current supply by battery B1=I1 = 2.56A (discharge)

Current supply by battery B2=I1 − I2 = 2.56 − 1.82 = 0.74A(current)

Current supply by battery B3=I2 = 1.82A(discharge)

Current supply by battery B4=I2 − I3 = 1.82 + 3.13 = 4.95A(discharge)

Current supply by battery B5=I3 = −3.13A(discharge)

Q.No.9.Solve the given circuit by nodal

analysis method to find the current in

various resistor of the circuit shown in

fig.

soln

The given circuit can be redrawn as in fig

below.

At node 1

V1 (1

2+

1

2+

1

10) −

V2

2−

V3

10= 28

∴ 11V1 − 5V2 − V3 = 280……………(i)

At node 2

V2 (1

2+

1

5+ 1) −

V1

2−

V3

1= 0






∴ 5V1 − 17V2 + 10V3 = 0……………(ii)

At node 3

V3 (1

4+ 1 +

1

10) −

V2

1−

V1

10= 0

∴ V1 − 10V2 − 12.5V3 = 20 …………………………(iii)

Solving equation (i),(ii) &(iii)

V1=37.32 V2=22.277 V3=19.20

Now,

I1=V1

2=

37.32

3=18.66A

I2=V1−V2

2=

37.32−22.277

2= 7.52A

I3=V1−V3

10=

37.32−19.2

10 = 1.812A

I4=V2−V3

1=

22.277−19.20

1= 3.077A

I5=V2

5=

22.277

5= 4.45A

Q.No.10. Using nodal analysis find the

different branch current in the given circuit.

All branch conductance are in siemen(i.e

mho). Also redraw circuit with values of

current.

soln

In frist node

V1(1 + 2) − V2X1 − V3X2 = −2

∴ 3V1 − V2 − 2V3 = −2………(i)

In second node

V2(1+4) – V1x1=5

∴ V1 – 5V2= – 5………(ii)

In third node

V3(2+3) – V1x2= – 5

∴ 2V1 – 5V3= 5……………(iii)

Solving equation (i),(ii) & (iii) we get

V1= – 3

2 = – 1.5

V2= 7

10 = 0.7

V3= – 8

5 = – 1.6

I1= (V1 – V2)X1=( – 1.5 – 0.7)x1 = – 2.2A

I2= (V3 – V1)X2=( – 1.6+1.5)x2= – 0.2A

I4= V2X4=0.7X4=2.8A

I3= (2A+ I4)=2A+2.8A=4.8A

Q.No.11.find the branch current in the circuit of given fig by using

a.Nodal analysis b.Loop analysis(mesh analysis)






soln

let

R1=6 Ω R2=2 Ω R3=4 Ω R4=3 Ω R5=4 Ω

E1=6V E2=10V E3=5V

a. Nodal analysis

For node A

VA (1

R1+

1

R2+

1

R3) −

E1

R1−

VB

R2+

E3

R2= 0

VA (1

6+

1

2+

1

4) −

6

6−

VB

2+

5

2= 0

∴ 2VA – VB= – 3…………(i)

For node B

VB (1

R5+

1

R2+

1

R3) −

E2

R3−

VA

R2−

E3

R2= 0

VA (1

4+

1

2+

1

4) −

10

4−

VA

2+

5

2= 0

∴ VA – 2VB= – 10…………(ii)

Solving equation (i) & (ii) we get

VA= 4

3V VB =

17

3V

I1= 𝐸1−𝑉𝐴

𝑅1=

6−(4

3)

6=

𝟕

𝟗A

I2= 𝑉𝐴+𝐸3−𝑉𝐵

𝑅2=

(4

3)+5−(

17

3)

2=

𝟏

𝟑A

I3=𝐸2−𝑉𝐵

𝑅3=

10−(17

3)

4=

𝟏𝟑

𝟏𝟐A

I4=𝑉𝐴

𝑅4=

(4

3)

3=

𝟒

𝟗A

I5=𝑉𝐵

𝑅5=

(17

3)

4=

𝟏𝟕

𝟏𝟐 A

b. Loop analysis

In loop A

– 6I1 – 3(I1 – I2)+6=0






∴ 3I1 – I2=2 …………….(a)

In loop B

5 – 2I1 – 4(I2 – I3) – 3(I2 – I1)=0

∴ 3I1 – 9I2+4I3= – 5……….(b)

In loop C

– 4I3 – 10 – 4(I3 – I2)=0

∴ 2I2 – 4I3=5………….(c)

Solving eqution (a),(b)&(c) we get

I1= 7

9A I2=

1

3A I3=−

13

12A

The – ve sign of I3 shows that the direction of current is opposite to that shown in fig.

Now the currents at various branch are as follow

Iab=I1= 7

9A

Ibf=I1 – I2= 7

9−

1

3=

4

9A

Ibc= I2= 1

3A

Ice=I2 – I3=1

3+

13

12=

17

12A

Idc= I3=−13

12A

Q.No.12.Three capacitor A, B &C have capacitance 10,50 & 25μF respectively.

Calculate

a) charge on each capacitance when connected in parallel to a 250V supply

b) Total capacitance when connected in parallel

c) P.D across each when connected in series

soln,

a.Q1=C1V

0r Q1=10x250

∴Q1=2500μf

Q2=C2V=50x250

0r, Q2=50x250

∴Q2=12500μf

Q3=C3V

0r, Q3=25x250

∴ Q3=6250μf

b. C= C1+ C2+ C3=10+50+25=85μf

c.1

c=

1

10+

1

50+

1

25

1

c=

25x50+10x25+50x10

10x50x25

C=6.25 μf

Then, total Q=CV=250x6.25=1562.5 μc

V1 =Q

C1=

1562.5

10= 𝟏𝟓𝟔. 𝟐𝟓𝐕

V2 =Q

C2=

1562.5

50= 𝟑𝟏. 𝟐𝟓𝐕

V1 =Q

C1=

1562.5

25= 𝟔𝟐. 𝟓𝐕






Q.No.13.Find the charges in the capacitor shown below and P.d across them.

soln

Ceqv= 2x(3+5)

(2+(3+5))= 1.6

Now,

Total charge Q=CV=1.6x100=160 μc

V1 =Q

C1=

160

2= 80V

V2 = 100 − 80 = 20V = V3

Q1 = C1V1 = 2x80 = 160μc

Q2 = C2V2 = 3x20 = 60μc

Q3 = C3V3 = 5x20 = 100μc

Q.No.14. Find Ceqv of the given circuit

Soln

CBCD= 1𝑥2

1+2=

2

3 μF

CBE=4+2

3=

14

3μF

CABE=2𝑥

14

3

2+14

3

=1.4 μF

Ceqv=CAF= 3+1.4=4.4 μF

Q.No.15. Two capacitor A & B are connected in series across a 100V supply and it is

observed that the p.d.s across them are 60V and 40V respectively. A capacitor of 2μF

is now connected in parallel with A and pd across B rises to 90V. Calculate the

capacitance of A and B.

soln

As capacitor A & B are

connected in series charge

across them is same so

CAVA= CBVB 60CA=40CB

∴ CA =2

3CB………………(1)

After 2 μF capacitance

connected across capacitor A.

combine capacitor on parallel

is

(CA+2) μF

Again charge is same on series so

(CA+2)x 10=90CB

or (2

3CB+2)= 9 CB

∴ CB=0.24 μF

∴ CA =2

3CB =

2

3x0.24 =0.16 μF

Q.No.16.Two coils with a coeffcient of coupling of 0.5 between them are connected in

series so as to magnetize (a) in same direction (b) in different direction. The






corresponding values of total inductances are for a) 1.9H and for (b) 0.7H. find the

self inductance of two coils and the mutual inductance between them.

soln

let mutual inductances be L1 & L2

for Case a) L1+L2+2M=L=1.9……..(i)

for Case b) L1+L2 – 2M=L=0.7……..(ii)

subtracting (ii) from (i) we get

M=0.3H

Putting this value in (i) we get

L1+L2+2x0.3=1.9

∴ L1+L2=1.3H ………….(iii)

We have M=k√L1L2

√L1L2 =𝑀

𝑘=

0.3

0.5= 0.6

∴ L1L2=0.36

From (iii) we get (L1 + L1)2 − 4L1L2 =

(L1 − L1)2

1.324𝑋0.36 = (L1 − L1)2

L1 − L1 =0.5……………(iv)

Solving (iii)&(iv)

L1= 0.9H L2=0.4H

Q.No.17.The combined inductance of two coils connected in series 0.6H or 0.1H

depending on the relative directions of the currents in the coils. If one of the coils

when isolated has a self inductance of 0.2 H calculate mutual inductance and

coupling co-efficient.

soln

L1+L2+2M=L=0.6……..(i)

L1+L2 – 2M=L=0.1……..(ii)

subtracting (ii) from (i) we get

M=0.125H

Let L1=0.2H then substituting value of M & L1 in equation (i) we get

L2= 0.15H

Coupling co-efficient= 𝑀

√L1L2=

0.125

√0.2𝑥0.15= 0.72

∴ k=0.72

Q.No.18.Two coils of inductances 4H & 6H are connected in parallel. If their mutual

inductances is 3H, calculate the equivalent inductance of combination if

i) mutual inductance assists the self – inductance

ii) mutual inductances opposes the self – inductances.

soln

i)L=(L1L2−M2)

(L1+L2 – 2M)=

4𝑥6−32

4+6−2𝑥3= 3.75H

ii)L=(L1L2−M2)

(L1+L2+ 2M)=

4𝑥6−32

4+6+2𝑥3= 0.94H






Chapter three

Network theorems Superposition Theorem

“The current at any point in the linear circuit containing ore than one independent

source can be obtained by superimposing the currents at that point caused by each source

acting alone”

According to this theorem, if there are a number of e.m.fs. acting simultaneously in

any linear bilateral network, then each e.m.f. acts independently of the others i.e. as if the

other e.m.fs. did not exist. The value of current in any conductor is the algebraic sum of

the currents due to each e.m.f. Similarly, voltage across any conductor is the algebraic sum

of the voltages which each e.m.f would have produced while acting singly. In other words,

current in or voltage across, any conductor of the network is obtained by super imposing

the currents and voltages due to each e.m.f. in the network.

Process to use superposition theorem in Numerical

Q. find I in AB

Let E2 be zero, then circuit become as shown in fig:a

RAB=R2 // R3= R2 .R3

R2+ R3

Total resistance = R1+RAB

Current due to E1, I1 =E1

R1+RAB

We know

IxR2=IyR3 → Iy = IxR2

R3

Ix+Iy = I1 →Ix+IxR2

R3= I1→Ix(

R3+R2

R3)= I1

∴Ix= I1R3

R3+R2

Let E1 be zero, then circuit become as shown in fig:b

RAB=R2 // R1= R2 .R1

R2+ R1

Total resistance = R3+RAB

Current due to E2, I2 = E2

R3+RAB

We know

Ix’R2=Iy’R3 → Iy

’= Ix’R2

R3

or, Ix’+Iy

’= I2→Ix’+

I’xR2

R3= I2→Ix

’(R1+R2

R3)= I2






∴Ix’=

I2R1

R1+R2

Hence current in AB Is

I=Ix +Ix’


Q.No.1:By using Superposition Theorem.find the

current in resistance R shown in Fig:1 R1=0.005 Ω,

R2=0.004 Ω, R =1 Ω, EI=2.05 V, E2= 2.15 V

.Internal resistances of cells are negligible.

Solution

Removing E2 the circuit becomes

Resistances of 1Ω and 0.04 Ω are in parallel across poins A and C.

RAC=1//0.04 =1x0.04

1.04=0.038 Ω.

This resistance is in series with 0.05 Ω.

Hence, total resistance offered to battery E1=0.05+ 0.038 =0.088 Ω.

I= 2.05

0.088 = 23.3A.

Current through l Ω resistance,

I1=23.3 x 0.04

1.04 =0.896A from C to A.

When E1 is removed, circuit becomes

Combined resistance of paths CBA and CDA i.e

Reqv of AC =1 //0.05 ==1x0.05

1.05=0.048 Ω

Total resistance offered to E2 is =0.04 +0.048 =0.088 Ω.

Current I= 2.15

0.088 = 24.4 A.

Again, I2=24.4 x 0.05

1.05 = 1.16 A. from C to A

current through 1- Ω resistance when both batteries are present

I= I1+ I2=0.896 + 1.16 =2.056 A.






Q.No.2:Use Superposition theorem to find current I in

the circuit shown in Fig:b. All resistances are in ohms

Solution.

Replacing voltage source by a short and the 40 A current

sources by an open. And considering only 120A current

source circuit becomes as shown in Fig.1

Using the current-divider rule, we get

I1=120 x 50

150+50 =30 A.

Replacing voltage source by a short and the 120 A current sources by an open. And

considering only 40A current source circuit becomes as shown in Fig.2

Again, using current-divider rule

I2=40 x 150

150+50 =30 A.

Again

Considering only voltage source circuit becomes as shown in Fig.3. Using Ohm's.law,

I3=10

150+50 =0.05 A.

Since I1 and I2 cancel out due to equal and opposite direction,

I=I3= 0.05 A.

Q.No.3: Compute the power dissipated in the 9-Ω resistor of given figure by

applying the Superposition principle. The

voltage and current sources should be treated

as ideal sources. All resistances are in ohms.

Solution.

An ideal constant-voltage sources has zero

internal resistances whereas a constant-current

source has an infinite internal resistance.

When Voltage Source Acts Alone

This case is shown in Fig. 1 where constant-current source has been replaced by an open-

circuit i.e. infinite resistance .

Further circuit simplification leads to the fact that total resistances offered to voltage

source is






Rtotal =4 + (12//15) =4+ 12x15

12+15 =

32

3Ωas shown in Fig. 4

Hence current Itotal =3232

3

= 3

In fig 2 current is divided into two part. The part going alone AB is the one that also passes

through 9 Ω resistor. Hence current through 9Ω resistor is

r = 12x3

12+15= 4/3 A

When Current Source Acts Alone

As shown in Fig:a,

the voltage source is

replaced by a short-circuit

Further simplification gives

the circuit of Fig b. here 12Ω

and 4Ω are parallel so they

are replaced by their

eqivalent resistor 4x12

4+12 =3Ω

In fig:b, 4A current divides into two parts. By current divider rule . current through 9Ω

resistor is

r’=(3+6)x4

3+6+9 =2A

Since both r and r’flowin the same direction, total current through 9-Ω resistor is

I=r +r' = (4/3)+ 2 = (10/3)A

Power dissipated in 9 Ω resistor is

P=I2R=(10/3)2 x 9 =100 W

Q.No.4: Using Superposition theorem, find the value

of the output voltage Vo in the circuit of given

Fig:4.All resistances are in ohms.

Solution

(a) Replacing 4 A source by an open circuit and 6V

source by a short circuit , given circuit changes as in

fig:a.

Using the current-divider rule, we can find current i1 through the 2 Ω resistor

I1=6 x 1

(l + 2 + 3) = 1A

∴V0l= I1 x R2=1 x 2 =2 V.






(b)Replacing 6A sources by an open-circuit and 6 Vsource by a short circuit, given circuit

changes as in fig:b. The current i2 can again be found with the help of current-divider rule

I2= 4x(2+1)

(2+3+1)=

12

6= 2A

∴ V02= I2 x R2 =2 x 2 = 4 V

(c)Replacing 6A sources by an open-circuit and 4A source by a open circuit, given circuit

changes as in fig:c.

Voltage drop overR2= 2 Ω resistor =VXR2

1+2+3=

6X2

6 = 2V

The potential of point B with respect to point A is= 6-2 = +4V. Hence. V03=- 4 V.

According to Superposition theorem, we have

V0= V01 + V02+ V03= 2 + 4-4 = 2 V

Q.No.5: Use superposition theorem to determine the voltage v in the network of

given fig.A.

soln

In fig A there

is one

dependent

source v

3

which cannot

be killed or

deactivate

during the

process of

using

superposition

theorem.

Let v1 be voltage across 3 Ω resistor due to 30V source only. So let 5A current source

replace by open circuit & 20V source by short circuit as in fig:B

Applying KCL to node 1

30−v1

6−

v1

3+

v13

−v1

2= 0

∴ v1=6V

Now , let 30V & 20V source replace by short circuit as in fig:C to find v2 . applying KCL

to node 1.

v2

6− 5 −

v2

3+

v23

−v2

2= 0

∴ v2= – 6V

Again let 5A current source replace by open circuit & 30V source by short circuit as in

fig:D to find v3. Applying KVL

– 2i – 20 – 2i – 1

3( – 2i)=0

∴ i=6A

Hence according to ohm’s law the component of v that correspond to 20V source is

v3=2x6=12V

∴v=v1+v2+v3=6 – 6+12=12V






Q.No.6:Using the superposition theorem find the magnetude and direction of current

through each resistor in given circuit of fig:a

Let us short circuit 24V source then given circuit becomes like circuit as shown in fig:b.

then

Reqv = 3 + (12||6) = 3 +12x6

12+6= 7 Ω

Current through whole circuit (I)= V

R=

18

7A

Current through 3 Ω resistor = IBC=I= 18

7A

Current through 12 Ω resistor = ICA= Ix6

12+6=

18

7X

6

12+6=

6

7A

Current through 6 Ω resistor = ICD= Ix12

12+6=

18

7X

12

12+6=

12

7A

Let us short circuit 18V source then given circuit becomes like circuit as shown in fig:c

Reqv = 12 + (3||6) = 12 +3x6

3+6= 14 Ω

Current through whole circuit(I)= V

R=

24

14=

12

7A

Current through 12 Ω resistor = IAC=I= 12

7A

Current through 3 Ω resistor = ICB= Ix6

3+6=

12

7X

6

3+6=

8

7A

Current through 6 Ω resistor = ICD= Ix12

3+6=

12

7X

3

3+6=

4

7A

Now current due to both source is

Current through 12 Ω resistor = IAC – ICA = 12

7−

6

7=

6

7A(Direction A – C)

Current through 3 Ω resistor = IBC – ICB = 18

7−

8

7=

10

7A(Direction B – C)

Current through 6 Ω resistor = ICD + ICD = 12

7+

4

7=

16

7A(Direction C – D)

Q.No.7:Using the superposition theorem to find the current in R=8 Ω resistance in

branch AB for the circuit shown in fig:A.

soln for fig:A

Reqv I – A= 48||48= 48x48

48+48= 24 Ω






Reqv E – F= 5||(8+12)=5||20=5x20

5+20= 4 Ω

Now, equivalant circuit of given circuit becomes like circuit as shown in fig:B

Let us short circuit 28V source then the circuit becomes as shown in fig:C

Now,

Reqv= 4 + 4||(6 + 8||24) = 4 + 4|| (6 +8x24

8+24) = 4 + 4||12 = 4 +

4x12

4+12= 7 Ω

I = V

R=

14

7= 2A

By applying current divider rule we get

Current across CA ICA= Ix4

4+(6+8||24)=

2x4

4+(6+8x24

8+24)=0.5A

Current across AB (IAB)= ICAx24

24+8= 0.5 x

24

24+8=

3

8A

Let us short circuit 14V source then the circuit becomes as shown in fig:D

Now,

Reqv= 24 + 8||(6 + 4||4) = 24 + 8|| (6 +4x4

4+4) = 24 + 8||8 = 24 +

4x4

4+4= 28 Ω

I = V

R=

28

28= 1A

By applying current divider rule we get

Current across AC IAC= Ix8

8+(6+4||4)=

1x8

8+(6+4x4

4+4)=

8

16=0.5A

Current across AB (IAB)= I – ICA=1 – 0.5=0.5A

Hence total current in 8 Ω resistor in branch AB is IAB+IAB= 3

8+ 0.5 = 0.875A

Q.No.8:Calculate the voltage drop across the 3 Ω resistor of fig:A using superposition

principle.

Let us make both 15A sourse open circuit to find voltage due to 20V only as shown in

fig:B

Reqv= 6||(1+2)+3= 6𝑥(1+2)

6+1+2+ 3 = 5 Ω

I= 𝑉

𝑅𝑒𝑞𝑣=

20

5= 4A






∴ current on 3 Ω resistor = IBC=I=4A

∴ voltage across 3 Ω resistor(VBC)= 3x IBC=3x4=12V

Now, make one of 15A source open circuit and 20V source short circuit as in fig:C

The circuit can be re-arranged as in fig:D. now,

Reqv= 1||(6||3)+2=1||(6𝑥3

6+3+ 2) =1||4 =

4𝑥1

4+1=

4

5Ω

V= IReqv=15x4

5 =12V

As we know voltage is divided in series but remain same in parallel, by applying voltage

divider rule we get

Voltage at 2 Ω resistor =V2Ω = 𝑉𝑥2

2+3||6=

12x2

2+(3x6)

3+6

=6V

∴ voltage across 3 Ω resistor (VCB) = V – V2Ω =12 – 6=6V

Now, make another 15A source open circuit and 20V source short circuit as in fig:E

The circuit can be re-arranged as in fig:F. now,

Reqv= 2||(6||3)+1=2||(6𝑥3

6+3+ 1) =2||3 =

2𝑥3

2+3=

6

5Ω

V= IReqv=15x6

5 =18V

As we know voltage is divided in series but remain same in parallel, by applying voltage

divider rule we get

Voltage at 2 Ω resistor =V1Ω = 𝑉𝑥2

1+3||6=

18x1

1+(3x6)

3+6

=6V

∴ voltage across 3 Ω resistor (VBC)= V – V2Ω =18 – 6=12V

Hence net voltage on 3 Ω resistor = VBC – VCB+ VBC=12 – 6+12=18V

Q.No.9: For the circuit given in fig :A

a)determine current I1,I2,I3 when switch S is in position b

b)using the results of part a), determine the same current with switch ZS in position

a.






a)When s is on b then circuit becomes as in fig:B. which can be redraw as in fig:C

As there’re more than two source let us use superposition theorem.

Frist let us find current due to 130 V source

only by short circuting other voltage

sources as in fig:E

Reqv=2||4+2=2𝑥4

2+4+ 2 =

10

3 Ω

Toal current due to 130V

I130=13010

3

= 39A

∴ I1=39A

∴ I2= Ix4

2+4=39𝑥

4

6= 26A

∴ I3=I – I2=39 – 26=13A

Secondly let us find current due to 120 V

source only by short circuting other voltage

sources as in fig:F

Reqv=2+2||4=2+2𝑥4

2+4=

10

3 Ω


I120=12010

3

= 36A

∴ I2=36A

∴ I1= Ix4

2+4=39𝑥

4

6= 24A

∴ I3=I2 – I=36 – 24=12A

As 20V is open circuted it doesnot play any role.

Now according to superposition theorem current due to both sources are as follow

I1=39 – 24=15A

I2=36 – 26=10A

I3=10+15=25A

b) When s is on b then circuit becomes as in fig:C. which can be redraw as in fig:G.

now let us find current due to 20V source only by short circuting other voltage sources. As

in fig H

Reqv=2+2||4=2+2𝑥4

2+4=

10

3 Ω


I120=2010

3

= 36A

∴ I2=6A ∴ I1= Ix4

2+4=6𝑥

4

6= 4A ∴ I3=I2 – I=6 – 4=2A

Hence net current I1,I2 & I3 are as follow

I1=15 – 4=11A

I2=10+6=16A

I3=25+2=27A






Thevenin’s Theorem

It provides a mathematical technique for replacing a given network, as viewed

from two output terminals, by a single voltage source with a series resistance. It makes the

solution of complicated networks (particularly, electronic networks) quite quick and easy.

Statement

Any two points across a resistance of network can be replaced by an equivalent

voltage source together with an equivalent series resistance. The equivalent voltage

source is equal to open circuit voltage across two points & the equivalent series

resistance is equal to equivalent resistance across two point when looking back into the

network from two points with all source replace by their internal resistances.

In other word “the current IL flowing through a resistance RL connected across any two

terminals of the metwork containing one or more voltage source is given by

IL =Vth

Rth + RL

Where Vth= opern circuit voltage between two terminals with RL connected

Rth=equivalent resistance of network with RL disconnected and the source of

current/voltage replaced with their internal resistance if any. ”

How to Thevenize a Given Circuit?

Steps to thevenize a given circuit.

1. Temporarily remove the resistance (called load resistance RL whose current is required)

2. Find the open-circuit voltage Voc which appears across the two terminals from where

resistance has been removed. It is also called Thevenin voltage Vth.

3. Compute the resistance of the whose network as looked into from these two terminals

afterall voltage sources have been removed leaving behind their internal resistances (if

any) andcurrent sources have been replaced by open-circuit i.e. infinite resistance. It is

also called Thevenin resistance Rth

4. Replace the entire network by a single Thevenin source, whose voltage is Vthand

whose internal resistance is Rth.

5. Connect RLback to its terminals from where it was previously removed.

6. Finally, calculate the current flowing through RLby using the equation.

I = Vth

RL+Rth

Numerical process

For above fig:1

1.calculation of Vth

Removing load resistance RL given circuit became as fig:2






Vth=VAB = R2

R2+r+R1X E where E=Vs

2. calculation of Rth

Short circuting voltage source(open circuit if current source is

given) the given Circuit became as fig:3

Rth=(R2//R1 + r) =R2XR1+r

R2+R1+r

3.Calculation of IL

Now replacing entire given network(circuit) by a single thevelin

source whose voltage is Vth and resistance is Rth, the thevenin’s equivalent circuit becomes

as shown in fig: 4

Now

IL = Vth

RL+Rth


Q.No.1:With reference to the network of Fig:a by applying Thevenin's theorem find

the following:

(i) The equivalent e.m.f of the network when viewed from terminals A and B.

(ii) The equivalent resistance of the network when looked into from terminals A

and.B.

(iii) Current in the load resistance RL of 15

Solution

(i) Current in the network before load resistance is connected [Fig.a]

I=V

∑R =

24

12+3+1 =1.5 A

:. voltage across terminals AB , Voc=Vth=IR= 1.5x 12 =18 V

Hence, so far as terminals A and B are concerned, the network has an e.m.f. of 18 volt

(and not 24 V).






(ii)There are two parallel paths between points A and B. Imagine that battery of 24 V is

removed but not its internal resistance. Then, resistance of the circuit as looked into

from point A and B is Fig.b

Rth =12//(3+1)=12//4=12X4

12+4=3 Ω

(iii) When load resistance of 15 Ω is connected across the terminals, the network is

reduced tothe thevenin’s equivalent circuitas shown in Fig. c

I = Vth

RL+Rth=

18

15+3= 1A

No.2:Using Thevenin theorem, calculate the current

flowing through the 4 Q resistor of given Fig.

Solution.

(i) Finding Vth

If we remove the 4Ω resistor, the circuit becomes as

shown in Fig.a Since full 10 A current passes through 2Ω

resistor, voltage drop across it is

V=IR=10 x 2 =20 V.

Hence, VB=20 V

The two resistors of 3Ω and 6Ω are connected in series across the 12 V battery. Hence,

drop across 6Ω resistor =12 x 6/(3 + 6) =8 V. ∵ V6=V x R6/(R6+R3)

∴ VA = 8 V

∴ Vth = VBA =VB- VA =20 - 8 =12 V-with B at a higher potential

(ii) Finding Rth

Now, we will find Rth i.e. equivalent resistnace of the network as looked back into the

open circuited terminals A and B. For this

purpose, we will replace both the voltage by a

short circuit and current sources replaced by

an 'open' i.e. infinite resistance. In that case,

the circuit becomes as shown in Fig:b which

is equivalent to fig:c

Rth= 6//3 + 2 = 4Ω. Hence,

Thevenin's equivalent circuit consists of a

voltage source of12V and a series resistance

of 4Ω as shown in Fig:d

When 4Ω resistor is connected across

terminals A and B, as

shown in Fig:e

I= 12/(4 + 4) = 1.5 A-from B to A (∵ I=V/Reqv)






Q.No.3:Use Thevenin's theorem to find the current

flowing through the 6Ω resistor of the network

shown in Fig:3. All resistances are in ohms.

Solution

For finding Vth

When 6 Ω resistor is removed given circuit reduced

as shown in Fig:a

Whole of 2A current flows along 2Ω in DC

producing a voltage drop

V=IR=(2 x 2) =4 V

As we go along BDCA,

the total voltage = - 4 + 12

=8 V -with A positive w.r.t. B. (∵

12V & 4V are in opposite direction)

Hence, Vth= 8 V

For finding Rth,

12 V voltage source is replaced by a short-circuit and the 2A current source by an

open-circuit, as shown in Fig:b.

The two 4Ω resistors are in series and are thus equivalent to an 8 a

resistance. However, this 8 a resistor is in parallel with a short of 0Ω.

Hence, their equivalent value is 0Ω. Now this 0Ω resistance is in series with

the 2Ω resistor.

Hence, Rth=2 + 0 =2Ω.

The Thevenin's equivalent circuit is shown in Fig:c

:. I = 8

2+6=1 Amp -from A to B

Q.No.4:The given circuit contains two voltage

sources and two current sources. Calculate

(a) Vth and (b) Rth between the open

terminals A and B of the circuit. All

resistance values are in ohms.

Solution.

For finding Vth

It should be understood that since terminals A and

B are open, 2 A current can flow






only through 4Ω and 10Ω resistors, thus

producing a drop of

V1=(2x10)=20 V across the 10Ω resistor, as

shown in Fig:a

Similarly,

3 A current can

flow through its own closed circuit between A and C

thereby producing a drop

V2= (8x3)=24 V across 8Ω resistor as shown in

Fig.a

Also, there is no drop across 2Ω resistor because no

current flows through it.

Starting from point B and going to point A via points D and C, we get

Vth=V1+V2-V= 20+ 24 - 20 =24 V -with point A positive.

For finding Rth

short-circuit the voltage sources and open-circuit the current sources, as shown in

Fig.b

Rth =RAB =8 + 10 + 2 =20 Ω.

Q.No.5: Calculate the power which would be

dissipated in the 8Ω resistor connected across

terminals A and B of given Fig. All resistance values

are in ohms.

Solution.

For finding Vth

The Thevenin's voltage V this equals the voltage drop

across 10Ω resistor between points C and The with

AB on open-circuit,

120-V battery voltage acts on the two parallel paths EF and ECDF.

Hence,

current through 10Ω resistor is

I10= 120

20+10+20 =2.4 A

Drop across l0Ω resistor,

Vth= 10 x 2.4 =24 V






For finding Vth

120 V battery is removed. It results in shorting the 40-Ω resistance since internal

resistance of the battery is zero as shown in Fig.a

Rth = 16+10//(20+20)=16+ 10x(20+20)

10+(20+20) +16=40Ω

Thevenin's equivalent circuit is shown in Fig:b

When 8-Ω resistoris connected in terminal AB as in fig:c then

Current through it is

I=24

40+8 =

1

2A

And power dissipated is

∴P = I2 R =(1

2)2x 8 = 2W

Q.No.6: Calculate the current in 8 Ω resistor of given fig:A by using thevenin’s

theorem.

soln

let us take out 8Ω resistor from the circuit then circuit became as shown in fig:B

As 6 Ω resistor is seen from AB the voltage of 6 Ω resistor is Vth

voltage on 6 Ω resistor due to 6V = 6x6

6+2+1=

6𝑥6

9= 4V

total voltage on 6 Ω resistor(Vth)= 12 –4= 8V

now for Rth let us make all voltage source short circuit as in fig:C then

Rth= 6||(2+1) = 6𝑥3

6+3=

18

9= 2 Ω

Hence current in 8 Ω resistor (I)= 𝑉𝑡ℎ

𝑅𝑡ℎ+𝑅𝐿=

8

2+8=

8

10= 0.8A

Q.No.7: Use thevenin’s theorem to calculate the p.d across terminals A & B shown in

fig:A below.

soln

let us take out 3Ω resistor on AB from the circuit then circuit became as shown in fig:B

To find Rth let us make all voltage source short circuit as in fig:C then

Rth= 3||6||6 = 3x6x6

3x6+6x6+6x3= 1.5 Ω






As 3 Ω resistor is seen from AB the voltage of 6 Ω resistor is Vth

As there are more than one voltage source it’ll be easy to use superposition theorem to

find voltage on 3 Ω resistor.

Frist let us take only 6V source on HG then circuit become as in fig:D

Reqv= 6+6||3=6+ 6x3

6+3= 8 Ω

I = V

Reqv=

6

8=

3

4A

Current on 3 Ω resistor I3=

3

4x6

3+6=

1

2A

Voltage on 3 Ω resistor Vdc= I3x3 =1

2x3 =

3

2 V

Again take only 6V source on EF then circuit become as in fig:E

Reqv= 6||3+6 = 6x3

6+3+ 6 = 8 Ω

I = V

Reqv=

6

8=

3

4A

Current on 3 Ω resistor I3=

3

4x6

3+6=

1

2A

Voltage on 3 Ω resistor Vcd= I3x3 = 1

2x3 =

3

2 V

Again take only 4.5V source on DC then circuit become as in fig:F

Reqv= 3+6||6=3+ 6x6

6+6= 6 Ω

I = V

Reqv=

4.5

6=

3

4A

Current on 3 Ω resistor I3= 3

4

Voltage on 3 Ω resistor Vdc= I3x3 = 3

4x3 =

9

4 V

Now Vth=VDC= VDC – VCD– VCD=3

2−

3

2−

9

4= −

9

4

∴ VCD= 9

4V

Now

Current through 3 Ω resistor on AB (I)= Vth

Rth+RL=

9

4

1.5+3=0.5A

Hence p.d across terminal A&B is given by

VAB= IR= 0.5 x 3 =1.5V

Q.No.8: Compute the current flowing through the load resistance of 10 Ω connected

across terminals A & B in given fig A by using thevenin’s theorem.






Frist load resistance 10 Ω is taken out from the given circuit as in fig:B. then to calculate

thevenin’s resistance shirt circuit the voltage source as in fig:C

Rth= 5+(10||10||10)+5= 10 +10𝑥10𝑥5

10𝑥5+5𝑥10+10𝑥10= 12.5 Ω

Now, for fig:B

VEF = 9 𝑥10

10+5= 6V

REF= 10||5 = 10𝑥5

10+5=

10

3 Ω

Now circuit changes to Fig:D

VCD = 6 𝑥10

10+10

3

= 4.5V

RCD= 10

3𝑥||10 =

10𝑥10

3

10+10

3

=5

2Ω

Again circuit changes to fig:E

According to thevenin’s theorem

Vth=Vcd=4.5V

Now

Current across 10 Ω load resistor = Vth

Rth+RL=

4.5

12.5+10=

1

5A = 0.2A

Q.No.9: Find thevenin’s equivalent circuit for terminal pair AB for the network

shown below in fig:A

soln






here 4A current only passes through 6 Ω resistor . So voltage on 6 Ω resistor is

VCA=IR=4x6=24V

20V only regulates on path cfgh so

Voltage on 10 Ω resistor (Vcf)= 20 x 10

10+15= 8V

∴ Vth=VAB= VAc+Vcf = – 24V+8V= – 16V i.e VBA=16V

to find Rth short circuit voltage source and open circuit current source then circuit network

changes as in fig:B

Rth=6+15||10+4= 10+ 15𝑥10

15+10=16 Ω

The thevenin’s equivalent circuit for given network for terminal points A & B is shown in

fig:C

Delta/Star Transformation

Sometimes experiences great difficulty in solving networks (having considerable

number of branches) due to a large number of simultaneous equations that have to be

solved. However, such complicated network can be simplified by successively replacing

delta meshes by equivalent star system and vice versa.

Delta connection to Star connection

Suppose three resistance R12 , R23 and R31 are connected in delta connection

between terminals 1, 2 and 3 as in Fig.a. This delta connected resistances can be replaced

by the three resistances R1 , R2 and R 3 connected in star as shown in Fig.b so that

resistance measured between any pair of terminals is same in both cases.

First, taking delta connection:

Between terminals 1 and 2, there are two parallel paths; one having a resistance of RI2 and

the other having a resistance of (R12+R31).

Resistance between terminals 1 and 2 is =(R12//R23+R31)=R12X(R23+R31)

R12+(R23+R31)…..…….(i)

Taking star connection:

The resistance between the same terminals 1 and 2 is (R1+ R2)……………………….(ii)

As terminal resistances have to be the same

Equating (i) & (ii)

R1+ R2=R12X(R23+R31)

R12+(R23+R31) ……………………….(iii)

Similarly, for terminals 2 and 3 we get

R2+ R3=R23X(R31+R12)

R23+(R31+R12)……………………….(iv)

Similarly, for terminals 3 and 1 we get,

R3+ R1=R31X(R12+R23)

R31+(R12+R23)……………………….(v)

Adding equation (iii),(iv)&(v)






R1+ R2+ R1+ R2+ R3+ R1= R12X(R23+R31)

R12+(R23+R31)+

R23X(R31+R12)

R23+(R31+R12)+

R31X(R12+R23)

R31+(R12+R23)

2(R1+ R2+ R3)= R12R23+R31R12

R12+R23+R31 +

R23R31+R23R12

R12+R23+R31 +

R31R12+R31R23

R12+R23+R31

2(R1+ R2+ R3) = 2(R12R23+R23R31+R31R12)

R12+R23+R31

∴R1+ R2+ R3 = R12R23+R23R31+R31R12

R12+R23+R31………………………….(vi)

Subtracting (iii) from (vi), we get

R3= R31R23

R12+R23+R31…………………………..………(1)

Subtracting (iv) from (vi), we get

R1= R31R12

R12+R23+R31…………………………………..(2)

Subtracting (v) from (vi), we get

R2= R23R12

R12+R23+R31 ……………..……………………(3)

Above equations are for converting Delta connection to Star connection. Resistance of

each arm of the star is given by the product of the resistances of the two delta sides that

meet at its end divided by the sum of the three delta resistances.

Star connection to Delta connection

This transformation can be easily done by using equations (iii), (iv) and (v) given

above. Multiplying (v) and (iv), (iv) and (v), (v) and (iii) and adding them together and

then simplifying them, weget

R12=R1R2+R2R3+R3R1

R3

R23= R1R2+R2R3+R3R1

R1

R31= R1R2+R2R3+R3R1

R2

The equivalent delta resistance between any two terminals is given by the sum of star

resistances between those terminals plus me product of these two star resitances divide by

the third star resistances.


Q.N.1.Calculate the current flowing through the 10Ω resistor of given Fig.

Solution.






Given circuit has two deltas i.e.ABC and DEF. We can

convert them into their equivalent stars as shown in

Fig.a

Each arm of the delta ABC hasa resistance of 12Ω and

their equivalent star resistance are given by

RAo=ROB=Roc= 12X12

12+12+12 = 4Ω so, and each arm of the

equivalent star has a resistance of 4Ω.

Similarly, each arm of the delta DEF has a resistance of

30Ω and the equivalent star has a resistance

R= 30x30

30+30+30= 10Ωper arm.

The total circuit resistance between A and F is

Reqv=4 + (4+34+10) //(4+10+10) + 10

=14+48//24

=14+48x24

48+24 = 30Ω.

Hence Itotal=V

R =

180

30 =6 A.

Current through 10 Ω resistor as given by current-divider rule is

I10=Itotal xRopposite

Rtotal =6l x

(4+34+10)

(4+34+10) + (10+10+4) =4 A.

Q.N.2.A bridge network ABCD has arms AB, BC, CD

and DA of resistances 1, 1, 2 and 1 Ω respectively. If

the detector AC has a resistance of 1 ohm, determine

by star/delta transformation, the network resistance

as viewed from the battery terminals.

Solution

Delta connection DAC(part of given figure) shown in

fig:a can be reduces to its equivalent star connection as

shown in fig.b by following way

RD= 1x2

1+2+1 = 0.5Ω

RA= 1x1

1+2+1 = 0.25Ω

Rc= 1x2

1+2+1 = 0.5Ω

Hence, given figureis reduced to the one shown in fig.c.

There are two parallel paths between points N and B. Their combined resistance is






RNB= (0.25+1)//(0.5+1)=(1.25)//(1.5) =1.25𝑋1.5

1.25+1.5=

1.875

2.75= 0.681818Ω

Reqv=RNB +0.5=0.681818Ω+0.5= 1.181818Ω

Q.N.3. Use delta-star conversion to find resistance between terminals 'AB' of the

circuit shown in Fig. 3 .All resistances are in ohms

Solution

First applying delta-star conversion in CGD loop

RC= 2X4

2+4+4=

8

10=0.8Ω

RD= 2X4

2+4+4 =

8

10= 0.8Ω

RG= 4X4

2+4+4 =

16

10=1.6Ω

Applying delta-star conversion in EGF loop

RE= 2X4

2+4+4 =

8

10 =0.8Ω

RF= 2X4

2+4+4 =

8

10 = 0.8Ω

RG= 4X4

2+4+4 =

16

10=1.6Ω

Now new circuit became as shown in fig:a which is Equivalent to another fig b

In fig:b 2Ω& 0.8Ω resistor are in series hence

RCJ=2Ω + 0.8Ω=2.8Ω

REJ=2Ω + 0.8Ω=2.8Ω

Fig:B Circuit became as shown in fig:c

Again We can change CHJ delta connection into Star

connection as follow

Rc=0.8X2.8

0.8+2.8+3.2 = 0.33Ω

RH=0.8X3.2

0.8+2.8+3.2 = 0.376Ω

RJ=3.2X2.8

0.8+2.8+3.2 = 1.32Ω

Circuit of fig:c changes to fig:d.

RAB=5+0.33+(1.32+2.8)//(0.378+0.8)

RAB=5.33+(4.12//1.178) =5.33+ 4.12X1.178

4.12+1.178=5.33 +

0.916

∴RAB=6.245






Q.N.4. A network of resistances is formed as in given Fig. AB = 9Ω; BC = 1Ω; CA =

1.5 Ω forming a delta and AD = 6 Ω ; BD = 4 Ω and CD =3 Ω forming a star.

Compute the network resistance measured between (i) A and

B (ii) Band C and (iii) C and A.

Solution

The star of given Fig. may be converted into the equivalent delta

and combined in parallel with the given delta ABC.

RAB1=

6x3+3x4+6x4

3= 18Ω

RBC1=

6x3+3x4+6x4

6= 9Ω

RCA1=

6x3+3x4+6x4

4= 13.5Ω

Then given circuit becames changes to fig:a

In fig:a 9Ω &18Ω are parallel in AB,1.5Ω & 13.5Ω are parallel in AC And 9Ω & 1Ω are

parallel in BC. Hence

RAB2=

9x18

9+18= 6Ω

RCA2=

1.5x13.5

1.5+13.5=

27

20Ω

RCB2=

9x1

9+1=

9

10Ω

Again fig:a changes to fig:b

Here

RAB= 6//(27

20+

9

10)=6//2.25=

6x2.25

6+2.25=

18

11 =1.6363Ω

RBC= 9

10//(6+

27

20)= 0.9//7.35=

0.9x7.35

0.9+7.35= 0.8018Ω

RCA=27

20//(6+

9

10)= 1.35//6.9=

1.35x6.9

1.35+6.9= 𝟏. 𝟏𝟐𝟗Ω

Q.N.5. Find the current in 17Ω resistor in the network shown in fig:A by using Delta-

star transformation.

soln

RAC=2+4=6 Ω REG=11+4=15 Ω

Now we can arrange fig:A as in fig:B

let us apply star-delta transformation on ACH & DEG

As RAC= RAH= RCH on ACH then their corresponding delta will also be equal

RA=RC=RH= 6𝑥6

6+6+6= 2 Ω

As RDE= REG= RGD on DEG then their corresponding delta will also be equal

RD=RG=RE= 15𝑥15

15+15+15= 5 Ω

Now circuit will be as in fig:C






Reqv= 2+(2+41+5)||(2+17+5)+5 = 7+48||24=7 +48𝑥24

48+24=23 Ω

I = 115

23= 5A

Current on 17 Ω resistor = I x 2+41+5

(2+41+5)+(2+17+5)=

5𝑥48

72=

10

3A

Q.N.6. find equivalent resistance between point A& B of given fig:A.

Soln

Given circuit can be re circuted as shown in fig:B. let us apply star-delta transformation in

ACD

RA=2RXR

2R+R+R= 0.5R

RC=2RXR

2R+R+R= 0.5R

RC=RXR

2R+R+R= 0.25R

Now circuit becameas shown in fig:C

Reqv=RAB=0.5R+(0.5R+R)||(0.25R+2R)=0.5R+ 1.5Rx2.25R

1.5R+2.25R= 1.4R

∴ Reqv=1.4R

Maximum power transfer theorem:

It states that “Maximum power is transferred from a source to load when the load

resistance is made equal to the internal resistance of the source as view from the load

terminals when load removed and all exit source are replaced by their interval resistance.”






In figure Ri is equivalent to thevnin’s

resistance (Rth) and V is equal to

thevnin’s voltage (Vth) so maximum

power is transferred from the circuit to

load when RL is made equal to Ri .

Proof:

Let us consider a circuit shown above

then circuit current is given by

I =V

Ri+RL

Power deliver to load P = I2RL = (V

Ri+RL)

2

RL…….(i)

The value of V and Ri for a given source is constant so power deliver to the load depends

upon RL. To find the value of RL for which maximum value of power is obtained, we can

differentiate equation (i) with respect to RL and set to result zero, then

dp

dRL=

d

dRL((

V

Ri+RL)

2

RL) = 0

V2 [((Ri+RL)2)−2RL(Ri+RL)

(Ri+RL)4] = 0

((Ri + RL)2) − 2RL(Ri + RL) = 0

(Ri + RL)(Ri + RL − 2RL) = 0

Since Ri + RL cannot be zero (Ri + RL − 2RL) = 0 (Ri − RL) = 0

Ri = RL

Proved.

Maximum power transferred(Pmax)=I2RL = (V

Ri+RL)2RL

Pmax = (V

RL+RL)2RL = (

V

2RL)2RL

∴ Pmax =V2

4RL

The voltage across load at maximum power is given by,

Load voltage=IxRL =V

Ri+RLxRL =

V

RL+RLxRL =

V

2RLxRL =

V

2

Q.No.1:Find the value of load resistance RL in given fig below

i)for transfer of maximum power

ii)determine also maximum power

Solution,

Calculation of Vth from fig:A

Vth=120x20

20+40= 40

Calculation of Rth

Making voltage source zero circuit becomes as in fig:B






Rth = 60 + 20||40 = 60 +20x40

20+40= 60 +

800

60= 73.33

For transfer of maximum power the value of load resistance is given by

RL=Ri=Rth=73.33Ω

The maximum power is given by

Pmax=V2

4RL =

402

4x73.33= 5.45watt

Q.No.2: Calculate the value of Are which will absorb maximum power from the

circuit of fig:A. Also compute the value of maximum power.

soln

Let us remove R and find thevenin’s voltage across A & B as shown in fig:B.

Now lwt us convert 120V source to current source. ∴ I=V

R=

120

10= 12A now circuit

become as in fig:C. by applying KCL we get Vth

10+

Vth

5= 12 + 6

∴ Vth=60V

Now to find Rth open circuit both current source in fig:C then circuit become as in fig:The

Rth= 10||5=10𝑥5

10+5=

10

3 Ω

The thevenins equivalent network is shown in fig:E

According to maximun power transfer theorem, R will absorb maximum power when it

equals to 10

3 Ω . In that case

I=V

Reqv=

6010

3+

10

3

= 9A

Maximum power (Pmax)=I2R=92x10

3=270W

∴ Pmax =270W






Chapter four Ac fundamentals

Generation of Alternating current and voltage

AC emf may be generated by rotating a coil in a magnetic field. as shown In

Fig.(a) or by rotating of magnetic field within a stationary coil, as shown in Fig.(b). The

value of emf is depends upon the angle between magnetic field and conductor.

In fig.b conductor AB was kept around rotor (magnetic poles). When rotor is

rotated in particular direction the magnetic flux produced by rotor poles cuts conductor

AB continuously. Hence emf will induced in conductor. At various position in rotor poles

the magnitude and direction of flux linkage with conductor AB are different and

accordingly magnitude and direction of emf induced will also change.

Figure besides shows waveform of emf

generated in coil AB. This waveform generater Is

sine wave And can be describe by following

equation

E=Emsinωt

where

ω=Angular velocity of motor

Em=Maximum value of emf

t= time period to generate a cycle of emf

we know,

frequency of emf generated ‘f’ is defined as number of cycle generated in one second

∴ f =1

t

Angular displacement in t second =ωt

or, 2π = ωt

∴ ω = 2π

t= 2πf

Hence E=Emsin(2πft)






Waveform and terms used in AC Phase :

Phase of an alternating waveform is defined as

the fraction of time period that has elapsed

since the waveform has passed through the

zero position. For example: the phase of

waveform at position A in figure beside is 𝑇

4

or 𝜋

2

Phase difference

In fig beside

ec leads ea by phase of 𝜋

2

eb lags ea by phase of 𝜋

2

If ea= Emsinωt then

ec= Emsin(ωt+ 𝜋

2)

eb= Emsin(ωt - π

2)

Amplitude

The maximum value, positive or negative, of an alternating quantity is known as its

amplitude.

Frequency

The number of cycle per second is called the frequency of the alternating quantity. lts unit

is hertz(Hz). It may be noted that the frequency is given by the reciprocal of the time

period of the alternating quantity.

f= 1

t or t=

1

f

Root-Mean-Square (R.M.S.) Value

The r.m.s. value of an alternaling current is given by that steady (d.c) current which when

flowing through a given time produces the same heat as produced by the alternating

current when flowing through the same circuit for the same time.

Consider an AC i=Imsinωt passing through resister of RΩ as in above fig.

the waveform of current is shown on above right side fig..

Since magnitude of current I is different at instant the heat generated is also

different at different instant. Let us divide time period T into n equal small

divisions/intervals so that magnitude of current during small time interval of t/n is nearly

constant.

Heat energy produced in 1st interval =i12R

T

n






Heat energy produced in 2nd interval = i22R

T

n

Heat energy produced in 3rd interval = i32R

T

n

.

.

Heat energy produced in 1st interval = in2R

T

n

Total heat energy produced in T sec =( i12+ i2

2+ i32+ i4

2+…….+ in2)R

T

n

Suppose a steady current of magnitude I produced the same amount of heat energy in time

T, then

I2RT=( i12+ i2

2+ i32+ i4

2+…….+ in2)R

T

n

Or,I=√i12 + i2

2 + i32 + i3

2 ………………+ in 2 x

1

n = RMS value of I

Mathematically

Irms=√1

2π∫ i2dωt

2π

0

=√1

2π∫ (Imsinωt)2dωt

2π

0

=√Im

2

2π[ωt −

sin2ωt

2]0

2π

=√Im2

2π[2π −

sin4π

2]

∴ Irms =Im

√2

Where Im= Em

R Similarly, Vrms=

Vm

√2

Average value

The average value of an AC is given by that steady current which transfer across

any circuit the same amount of charge as transferred by that AC during that time. The

charge transferred at any instant is proportional to current at that instant.

Iav =i1+i2+i3………+in

n

or, Iav = 1

π∫ idωt

π

0=

1

π∫ Imsinωtdωt

π

0

or , Iav = Im

π[−cosωt]0

π = 2Im

π

∴ Iav = 2Im

π

Similarly

Vav = 2Vm

π

Phasor representation of sinusoidal waveform

A sinsoidal signal V = Vmsinωt can be reprasented by phasor rotating in

anticlockwise direction with constant angular velocity of ω and with a magnetude equal to

Vm so that vertical component of that rotating phasor at any instant represents instaneous

value of signal at that instant.






Fig:b shows in waveform of signal V = Vmsinωt.. this signal can be represented by

rotating vector rotating in anticlockwise direction with angular velocity of ω & with a

magnetude equal to |OA|=|Vm|

At instant ωt=0, OA don’t have vertical component so that V=0 at ωt=0

After θ1 rotation the vertical component OA=Oasinθ which corresponds to instaneous

value of V at ωt1=θ1.

It is useful to note here that V=Vmsinωt=imaginary part of Vmejωt

ejωt=cosωt+jsinωt & j=√−1

∴ the equation V = Vmsinωt.. also can be written as

V = ImVmejcost.

Resistance exited by sinusoidal voltage(phasor representation)

Above fig:a shows an ac circuit with a pure resistance R exited by sinusoidal

voltage described by equation

V = Vmsinωt…………………………….. (i)

If I be the instanteneous value of current through circuit applying ohm’s law, we

can write,

i =V

R=

Vmsinωt

R = Imsinωt……………………………….(ii) where Im =

Vm

R

Comparing (i)&(ii)

We can say that I is also ac in nature and phase difference with V. The waveform

and phasor diagram of V and I is shown in fig:b and fig:c respectively.






Instantaneous power

P=Vi=Vmsinωt x Imsinωt ∴ P = Vm Im(sinωt) 2

Average power:

Pav =1

2π∫ VmImsin2ωtdωt

2π

0

Pav =VmIm2π

∫ sin2ωtdωt

2π

0

Pav =VmIm2π

∫(1 − cos2ωt)dωt

2π

0

Pav =VmIm2π

∫(1 − cos2ωt)dωt

2π

0

=VmIm

2=

Vm

√2xIm

√2= Vrms Irms

∴ Pav =VmIm

2= Vrms Irms

Inductance exited by sinusoidal voltage(phasor representation)

Above fig:a shows an ac circuit with a pure inductance L exited by sinusoidal

voltage described by equation

V = Vmsinωt If e be the instanteneous value of emf induced then, we can write,

e =Ldi

dt…………………(i)

from figure e=Vmsinωt ……………………………….(ii)

From (i)&(ii) Ldi

dt= Vmsinωt

or,i =Vm

L∫ sinωtdt

or, i =Vm

Lx(−

cosωt

ω)

∴ i = −Vm

Lω x sin(ωt − 90) ………(iii)

Lω = 2𝜋𝑓𝑙 =inductance resistance

-ve sign implies waveform starts from –ve direction.






Comparing equation (i) & (iii)

We can say that I is also ac

in nature and phase difference with

e.here I lags V by 90° The

waveform and phasor diagram of V

and I is shown in fig:b & c

Instantaneous power

P=Vi=Vmsinωt x Imsin (ωt − 90)

∴ P = −Vm Im

2cos2ωt

Average power:

Pav =1

2π∫ −

Vm Im2

cos2ωt

2π

0

= 0

Therefore pure inductance doesnot

consume power. The plot of instanteneous power versus ωt is shown in fig:b

Capacitor exited by sinusoidal voltage(phasor representation)

Above fig:a shows an ac circuit with a pure capacitor exited by sinusoidal voltage

described by equation

V = Vmsinωt The instanteneous value of voltage across the capacitor is given by

𝑉𝑐 =q

C …………………………….. (i)

At any instant,

q=VC

∴q=Vmsinωt.C

Differentiating both side with respect to time we get, dq

dt= C d

(Vmsinωt)

dt

or, dq

dt= CωVmcosωt

or,𝑖 =Vm1

ωC

sin (ωt + 90)

∴ i = Imsin (ωt + 90) ……….(ii)

Where Im =Vm

Xc






and Xc =1

ωC=

1

2𝜋𝑓𝑐 is known as capacitance reactance of the capacitor.

Comparing (i)&(ii)

We can say that i is also ac in nature and phase difference with e.here i leads V by

90° The waveform and phasor diagram of V and I is shown in fig:b & fig:c respectively

Instantaneous power

P=Vi=Vmsinωt x Imsin (ωt + 90)

∴ P = −Vm Im

2sin2ωt

Average power:

Pav =1

2π∫

Vm Im2

sin2ωt

2π

0

= 0

Therefore pure capacitor doesnot consume

power. The plot of instanteneous power

versus ωt is shown in fig:11.64

R-L series circuit exited by Sinusoidal voltage(phasor representation)

Above fig:a shows an AC circuit with a resistance connected in series with an

inductance & supplied by an AC emf.

Suppose

V=RMS value of applied voltage

VR=voltage across R

VL=voltage across L

R=Resistance in ohm

L=Inductance in henry

I=RMS value of current through circuit

XL=Induca

nce

reactance=

2πfL

Now,

voltage drop in resistance ‘R’ is VR = IR , there is no

difference between VR & I

Voltage drop in inductance ‘L’ is VL = IXL , I lags V by 90

From kirchoff voltage law

V = VR + VL






From phaser diagram

V = √VR2 + VL

2

Or, V = √I2R2 + I2XL2 = I√R2 + XL

2

Or,I =V

√R2+XL2

∴ I =V

Z where 𝑍 = √R2 + XL

2 known as impedence of circuit.

Here I lags V by same angle Ф

Where, tan∅ =VL

VR=

IXL

IR=

XL

R

∴ ∅ = tan−1(XL

R)

Instantaneous voltage and current

V = Vmsinωt

i = Imsin (ωt − ∅) where Im =Vm

Z

Waveform of V and I is shown in fig:c

Power in circuit

Let us divide current I into two

component in phasor diagram as

IcosФ=component of I in phase with V

IsinФ=component of I perpendicular to V

Now,

Active power(P)=V x IcosФ

=IZIR

z=I2R

It is responsible for producing heat.

Reactive power(s) =V x IsinФ

=IZI𝑋𝐿

𝑧= 𝐼2𝑋𝐿var

The product of V and total current I is called Volt

ampere of current & Is given by

𝑄 = 𝑉 𝑥 𝐼 = 𝐼2𝑧 = √𝑃2 + 𝑆2

It is also known as apparent power of the circuit.

Power factor

Among two component of power active power ‘p’ depends upon a factor cosФ

which is known as power factor of circuit.

If 𝑅 = 𝑋𝐿, ∅ = tan−1 (XL

R) = 45° & power factor=0.707

( Active power=reactive power)

If 𝑅 > 𝑋𝐿, ∅ = tan−1 (XL

R) < 45° & power factor>0.707

( Active power>reactive power)

If 𝑅 < 𝑋𝐿, ∅ = tan−1 (XL

R) > 45° & power factor<0.707

( Active power<reactive power)

If 𝑋𝐿 = 0, ∅ = tan−1 (XC

R) = 0° & power factor= 1(max)

Reactive power=0

If 𝑅 = 0, ∅ = tan−1 (XC

R) = 90° & power factor= 0(min)

Active power=0






R-C series circuit exited by Sinusoidal voltage(phasor representation)


capacitance & supplied by an AC emf.

Soppose


R=Resistance in ohm

VR=voltage across R

Vc=voltage across C

C=Capacitance in Farad


Xc=Capacitive reactance=1

2𝜋𝑓𝑐 in ohm

Now,

voltage drop in resistance ‘R’ is VR = IR , there is no difference between VR & I

Voltage drop in capacitor ‘C’ is VL = IXc , I leads V by 90

From kirchoff voltage law

V = VR + Vc

From phaser diagram

V = √VR2 + Vc

2

Or, V = √I2R2 + I2Xc2 = I√R2 + Xc

2

Or,I =V

√R2+Xc2

∴ I =V

Z where 𝑍 = √R2 + Xc

2 known as impedence of circuit.

From phasor diagram we can say that I leads V by

same angle Ф

Where, tan∅ =Vc

VR=

IXc

IR=

XC

R

∴ ∅ = tan−1(XC

R)

Instantaneous voltage and current

V = Vmsinωt

i = Imsin (ωt + ∅) where Im =Vm

Z

Waveform of V and I is given besides

Power in circuit

Let us divide current I into two component in phasor diagram as



Now,







=IZIR

z=I2R

It is responsible for producing heat.

Reactive power(s) =V x IsinФ var

=IZIXC

z= I2XCvar

The product of V and total current I is called Volt

ampere of current & Is given by

Q = V x I = √P2 + S2

It is also known as apparent power of the circuit.

Power factor

Among two component of power active power ‘p’ depends upon a factor cosФ

which is known as power factor of circuit.

If R = Xc ,then ∅ = tan−1 (XC

R) = 45° & power factor=0.707

( Active power=reactive power)

If R > XC , then ∅ = tan−1 (XC

R) > 45° & power factor>0.707

( Active power>reactive power)

If R < XC , then ∅ = tan−1 (XC

R) < 45° & power factor<0.707

( Active power<reactive power)

If XC = 0, ∅ = tan−1 (XC

R) = 0° & power factor= 1(max)

Reactive power=0

If R = 0, ∅ = tan−1 (XC

R) = 90° & power factor= 0(min)

Active power=0

R-L-C series circuit exited by Sinusoidal voltage(phasor representation)


capacitance and inductance & exited by an AC voltage source V = Vmsinωt .

Suppose


R=Resistance in ohm

VR=voltage across R

VL=voltage across L

VC=voltage across C

C=Capacitance in Farad


XL=Inducance reactance=2πfL


2πfc in ohm






Case 1

Let, XL >Xc i.e VL>VC

For those phasor diagram will be as shown in fig below

Here I lags V by angleФ

Where,

∅ = tan−1(XL−XC

R)

Total voltage Is given by

V2 = (VL − VC)2 + VR

2

Or,V = √VR2 + (VL − VC)

2

Or, V = √I2R2 + I2(XL − XC)2

Or,𝑉 = I√R2 + (XL − XC)2

Or,I =V

√R2+(XL−XC)2

∴ I =V

Z where Z =

√R2 + (XL − XC)2 known as Impedence of circuit.

Case 2

Let, XC >XL i.e VC>VL

For those phasor diagram will be as shown in fig

below

Here I leads V by angle Ф

Where,

Ф = tan−1(XC−XL

R)

Total voltage Is given by

V2 = (VC − VL)2 + VR

2

Or,V = √VR2 + (VC − VL)

2

Or, V = √I2R2 + I2(XC − XL)2 =

I√R2 + (XC − XL)2

Or,I =V

√R2+(XC−XL)2

∴ I =V

Z where Z = √R2 + (XC − XL)

2 known

as Impedence of circuit.

Power in circuit

Let us divide current I into two component in phasor diagram as



Now,


Reactive power(s) =V x IsinФ var

Summary of result of AC circuit

Types of impedance Value of impedance Phase

angle for

current

Power

factor

Resistance only R 0° 1

Inductance only ωL 90°lags 0






Capacitance only 𝟏

𝛚𝐂

90° leads 0

Resistance+inductance √𝐑𝟐 + (𝛚𝐋)𝟐 0°<Ф<90°

lag

1>p.f>0

lag

Resistance+capacitance

√𝐑𝟐 + (𝟏

𝛚𝐂)𝟐

0°<Ф<90°

lead

1>p.f>0

lead

Resistance+inductance+capacitance

√𝐑𝟐 + (𝛚𝐋 −𝟏

𝛚𝐂)𝟐

Or

√𝐑𝟐 + (𝟏

𝛚𝐂− 𝛚𝐋)

𝟐

Between

0° and

90°

lags

Or

lead

Between

0 and

unity

lags

Or

lead

AC parallel circuit

Above figure show a typical ac circuit. Path 1 has a resistance ‘R’ in series with an

inductance ‘L’ and Path 2 has a resistance ‘R’ in series with an capacitor ‘C’. Total

combination is supplies by an AC voltage source.

Suppose,

V=RMS value of applied volage

I1=RMS value of current through path-1

I2=RMS value of current through path-2

XL=Inducance reactance=2πfL


2πfc in ohm

Here,

In path-1 I1 lags V by Ф1,where Ф1 = Tan−1 (XL

R1)

In path-2 I2 leads V by Ф2,where Ф2 = Tan−1 (Xc

R2)

The magnitude of I1 & I2 are given by:

I1 =V

Z+∅1 where Z1=√R1

2 + XL2 and ∅1 = Tan−1 (

XL

R1)

I2 =V

Z+∅2 where Z1=√R1

2 + XC2 and ∅1 = Tan−1 (

Xc

R2)






Phasor diagram is given below

I1cosФ1 = component of I1 in phase with V= active component of I1

I1sinФ1 = component of I1 which lags V by 90= reactive component of I1

I1cosФ2 = component of I2 in phase with V= active component of I2

I1sinФ2 = component of I2 which leads V by 90°= reactive component of I2

Net active component IX = I1cosФ1 + I1cosФ2

Net reactive component IY = I1sinФ1 + I1sinФ2

Then total current I is given by Phasor sum of Ix & IY as shown in phasor diagram. The

magnitude of total current I is given by

I = √(I1cosФ1 + I1cosФ2)2 + (I1sinФ1 + I1sinФ2)

2 = √IX2 + IY

2

Here I lags V by an angle Ф1. Where Ф1 = Tan−1 (IY

IX)

Total active power(P)=V x IcosФ

Total reactive power(s) =V x IsinФ

Resonance in R-L-C series circuit

Above figure shows R-L-C series circuit supplied by an ac voltage source.

Suppose,

L=inductance in henry

C=capacitance in farad

ω= Angular frequency in radian/sec

The reactance of inductor and capacitor depends upon the frequency of applied voltage

and are given by Inductive reactance, XL = 2πfL = ωL

capacitive reactance , XC =1

2πfC=

1

ωCΩ ………………(i)

Now net reactance of circuit=XL − XC &

Total impedence Z = √R2 + (XL − XC)2

From equation (i) it is clear that if frequency decreases then inductive reactance decreases

and capacitive reactance increases and vice versa.

At particular frequency net reactance becomes zero i.e.






XL − XC = 0 & Z=R, power factor is unity,the current I is maximum and

I =V

R . Such condition is known as resonance in R-L-C circuit. The frequency at which

this condition occures is known as resonance frequency.

XL − XC = 0 → XL = XC → 2πf0L =1

2πf0C

∴ f0 =1

2π√LC

Above fig shows variation of XL, XC & Z with respect to

frequency f. At resonance frequency f0 net reactance

X=0

Figure given in right side shows variation of current in

circuit w.r.t frequency ‘f’. At resonance frequency ‘f0’

net reactance, x=0

i.e

XL = XC

At frequency f<f0 , XL > XC,net impedence is capacitive

& power factor is leading.

At frequency f>f0 , XL < XC,net impedence is inductive

& power factor is lagging.

f1 & f2 are two frequencies at which I =I0

√2

power at these two frequencies =(I0

√2)

2

R =1

2I02R=half the power at fo

These two frequencies f1 & f2 are known as half power frequency. The value of f1

& f2 are given below

f1 = f0 −R

4πL f2 = f0 +

R

4πL

The range of frequencies between f1 & f2 is known as bandwidth of circuit. This

indicates that the signal within these range of frequency will easily pass through this

circuit and signal of frequency below f1 & above f2 will be attenuated by this circuit






Resonance in R-L-C parallel circuit

Above figure shows R-L-C parallel circuit supplied by an ac voltage source.

Suppose,

L=inductance in henry

C=capacitance in farad

R=resistance in ohm

ω= Angular frequency in radian/sec

Impedence of path I, Z1 = R + j(XL) = R + j(2πfL)

Current through path I, IL =V

Z1 =

V

R+j(2πfL)

Impedence of path II, Z2 = −j(XC) = −j1

(2πfC)

Current through path II, IC =V

Z2 = −j

V

(2πfC)

IL lags V by Ф=tan−1(XC

R) and IC leads V by 90°

The total current is phasor sum of IL and IC

At lower frequency XL is low and Xc is high.

∴ IL > IC

At higher frequency XL is high and Xc is low.

∴ IL < IC

At particular frequency Ic = ILsinФL so that total current I is in phase with V resulting

unity power factor as in fig c shown above. This condition is known as resonance in

parallel ac circuit.

At this condition

Ic = ILsinФL






V

XC=

V

Zx

XL

Z → XLXC = Z2

or, 2πf0l x1

2πf0C= Z2 →

L

C= Z2 = R2 + XL

2

or,L

c= R2 + (2πf0l)

2 →L

C− R2 = (2πf0l)

2

or, √L

C− R2 = 2πf0l

∴ f0 =1

2π√

L

C− R2

Which is required equation for resonance frequency

Above figure shows resonance curve for parallel ac circuit. The current is minium at

resonance frequency ‘fo’ and equal to VR

Z2

Solved numerical problems

Q.No.1. A coil having Resistance of 4Ω and inductance of 9.55mH is connected in

series across a supply of 240V, 50Hz. Calcuate

a) Inductive reactance

b) Impedance

c) Phase angle

d) Total current

e) Power factor

f) Voltage across resistor

g) Voltage across inductor

h) Active power

i) Reactive power

j) Apparent power

k) Draw a phasor diagram

Soln,

Given

Resistance (R)=4Ω

Inductance (L)=9.55mh=9.55 x 10-3H

Voltage(V)=240V

Frequency (F)=50Hz

Now,






a) Inductive reactance(XL)=ωL=2πfL= 2πx50 x 9.55 x 10-3=3Ω

b) Impedance(Z)=√R2 + XL2=√42 + 32 = √25 = 5 Ω

c) Phase angle( Ф)=tan−1 (XL

R) = tan−1 (

3

4) = 36.86°

d) Total current ( I )=V

Z=

240

5= 48A

e) Power factor = cos Ф=cos36.86° =0.80010 (lagging)

f) Voltage across resistor (VR)=IR=48 x 4=192V

g) Voltage across inductor(VL)=IXL=48 x 3 = 144 V

h) Active power (P)=I2R=(48)2x4=9216watt

i) Reactive power(Q)= I2XL=(48)2x3=6912VAR

j) Apparent power (S)= V x I=240x48=11520 or,√P2 + Q2 = 11520 VA

k) A phasor diagram is

Q.No.2. For a circuit below calculate

a) Power factor

b) RMS value of current

c) Frequency of supply voltage

d) Active power and reactive power

e) Expression for instantaneous

current in circuit

Soln,given

V=400sin314t

Vm=400

ω =314

R=15 Ω

L=0.1H

Now,

XL=ωL=314 x 0.1=31.4 Ω

Z=√R2 + XL2 = √152 + 31.42 = 34.8 Ω

Ф = tan-1(XL

R) = tan−1 (

31.4

15) = 64.47°

a) Power factor (pf)= cosФ=cos64.47° =0.43






IM= Vm

Z=

400

34.8= 11.49A

b) RMS value of current(Irms) =Im

√2=

11.49

√2= 8.13A

We have

ω =2𝜋f

c) ∴ Frequency of supply voltage (f)=ω

2π=

314

2π= 50Hz

d) Active power(p)= I2R= (8.13)2 x 15 =991.4535watt

Reactive power (Q)= I2XL=(8.13)2x 31.4=8685.132VAR

e) i= IMsin( ωt – Ф)= 11.49sin( 314t – 64.47°)

Q.No.3.A coil takes a current of 6A when connected to a 24V d.c supply. To obtain

the same current with a 50-hz a.c supply the voltage required was 30V. calculate

inductance of coil and power factor of coil.

Soln

As coil offers resistance in direct voltage & impedance to Ac voltage

R=24

6= 4 Ω

Z=30/6=5 Ω

∴ XL=√𝑍2 − 𝑅2 = √52 − 42 = 3 Ω

Power factor= 𝑅

𝑍=

4

5= 0.8(lag)

Q.No.4. A capacitor is connected in series with a 40 Ω resistor across a supply of

frequency 60Hz. A current of 3A flows and the circuit impedence is 50 Ω .Calculate

a) The value of capacitance

b) Supply voltage

c) Phase angle

d) P.d across resistor and capacitor

e) Power factor

f) Active power ,reactive power and

apparent power

g) Draw the phasor diagram

soln, given

Resistance(R) =40 Ω

Current (I)=3A

Frequency(f)=60Hz

Impedence (Z)=50Ω

Now,

Z=√R2 + Xc2 ↔ Xc = √Z2 − R2 =

√502 − 402 = 30 Ω

Xc =1

ωc=

1

2πfc

∴ c =1

2πfXc=

1

2πx30x60= 88.42μf

a) ∴ The value of capacitance ( c)=88.42μf

b) Supply voltage(V)=IZ=3x50=150V

c) phase angle (Ф)=tan−1 (Xc

R) =

tan−1 (30

40) = 36.87°

d) p.d across resistor (VR)=IR=3x40=120V

p.d across capacitor (Vc)=IXc=3x30=90V

e) power factor = cos Ф =cos36.87°=0.8 (leading)

f) Active power(p)= I2R= (3)2 x 40 =360watt

Reactive power (Q)= I2XL=(3)2x 30=270VAR

A apparent power (S)= VI=150x3=450VA

g) The phasor diagram is shown on fig:1






Q.No.5. A resistance of 100 Ω , inductance of 100mH and capacitance of 100 μf are

connected in series and supplied by an ac voltage source of 200 volts. Calculate

a) Resonance frequency

b) Half power frequency

c) Current at resonance

d) Bandwidth

soln,given

R=100Ω

L=100mH=100 x 10-3H

C=100 μf =100 x 10-6f

V=200V

Now,

a) Resonance frequency (f0)=1

2π√fc

=1

2π√100x10−3x100x10−6 = 50Hz

b) Half power frequency

f1 = fo −R

4πL= 50 −

100

4πx100x10−3 = −29.577Hz

f2 = fo +R

4πL= 50 +

100

4πx100x10−3 = 129.577Hz

c) Current at resonance (Io)=V

R=

200

100= 2A

d) Bandwidth =f2-f1=129.577 − (−29.577) = 159.154Hz

Q.No.6. A coil with resistance of 50 Ω and inductance of 0.318H is connected in series

with a 159 μf capacitor. The resulting circuit is connected across a 220V,50Hz ac

supply. Find

a) P.f of coil and p.f of circuit

b) Circuit current

c) Active ,Reactive & Apparent power

d) Voltage across the coil and capacitor

Soln,

XL=ωL=2πfL= 2πx50x0.318=100 Ω

Xc =1

ωc=

1

2πfc=

1

2πx50x159x10−6 = 20 Ω

∴ Фcoil=tan−1 (XL

R) = tan−1 (

100

50) = 63.43

p.f of coil=cosФcoil=cos63.43=0.447

∴ Фcircuit=tan−1 (XL−Xc

R) = tan−1 (

80

50) = 58°

p.f of circuit=cosФcircuit=cos58=0.57

a) ∴ p.f of coil and circuit are 0.447 & 0.57 respectively

b)Circuit current (I)=V

Z=

V

√R2+(XL−Xc)2=

220

√502+(100−20)2= 2.33A

c)Active power (P)=I2R=(2.33)2x50=271.445watt

Reactive power (Q)=I2(XL-Xc)= (2.33)2x (100-20)=434.312VAR

Apparent power (s)=VxI=220x2.33=512.6VA

Now,

VR=IR=2.33x50=116.5V

VL=IXL=2.33x100=233V

VC =IXc=2.33x20=46.6V

d)Voltage across coil(Vcoil) = √VR2 + VL

2 = √116.52 + 2332 = 260.509V

Voltage across capacitor(Vcapacitor) =Vc=46.64 V






Q.No.7. Two impedance Z1=(10+j15) Ω & Z2 =(6-j8)Ω are connected in parallel.

The current supplied is 15A. what is the power taken by each branch?

soln

Z1=(10+j15) Ω =√102 + 152 = 18 Ω

Z2 =(6-j8)Ω = √62 + 82 = 10 Ω

∴ I1 =IxZ2

(Z1+Z2)= 15 x

10

18+10=

150

28= 5.357A

∴ I2 = IxZ1

(Z1+Z2)= 15 x

18

18+10=

270

28= 9.64 A

Power in branch 1 is P1

∴ P1= I12Z1 = 5.3572x18 = 516.55watt

Power in branch 2 is P2

∴ P2= I22Z2 = 9.642x10 = 929.296watt

Q.No.8. For given circuit, calculate

a) Power factor

b) Rms value of current

c) Impedence of the circuit

d) Active power and reactive power

e) Calculate the value of C for which p.f.=1

soln

XL=ωL=2πfL= 2πx50x100x 10-3=31.42 Ω

Xc =1

ωc=

1

2πfc=

1

2πx50x100x10−6 = 31.83 Ω

Impedence of whole circuit

Z =R + j XL − j Xc = 12 + j 31.42 − j 31.83 = 12 − j 0.41

Z=√122 + 0.412 = 12.007 Ω

The current through the circuit

I =V

Z=

200

12.007= 16.65 A

Now

a)power factor =cos Ф =cos1.957 (lead)

b)Irms=16.65A

c)Impedence =12.007

d) Active power =VI cosФ=200x16.65xcos1.957=3328.06 watt

Reactive power = VI sinФ=200x16.65xsin1.957=113.71VAR

e)For p.f=1

XL=XC

0r, XL= 1

2πfc

0r,C=1

2𝜋𝑓𝑋𝐿=

1

2𝜋𝑥50𝑥31.42= 101.32

∴ The value of C for p.f=1 is 101.32 μf

Q.No.9. For given circuit, calculate

a) Magnitude and phase of I1 and I2

b) Active power and reactive power of path-I

c) Active power and reactive power of path-II

d) Magnitude & power of total current I

e) Draw phasor diagram showing V,I,I1&I2

soln given,

R1=100 Ω L=100mH






R2=100 Ω C=200 μF V=200V f=50Hz

Now,

XL=ωL=2πfL= 2πx50x100x10-3=31.415 Ω

Xc =1

ωc=

1

2πfc=

1

2πx50x200x10−6 = 15.915 Ω

Z1= √𝑅12 + 𝑋𝐿

2 = √1002 + 31.4152 = 104.818 Ω

Z2= √𝑅22 + 𝑋𝑐

2 = √1002 + 15.9152 = 101.258 Ω

a)Magnitude of I1=𝑉

𝑍1=

200

104.818= 1.908A

Phase of I1= Ф1=tan−1 (𝑋𝐿

𝑅) = tan−1 (

31.415

100) = 17.30°

I1 lags V by Ф1=17.30°

Magnitude of I2=𝑉

𝑍2=

200

101.258= 1.975A

Phase of I2= Ф 2=tan−1 (𝑋𝑐

𝑅) = tan−1 (

15.915

100) = 9.042°

I2 leads V by Ф2,where Ф2 = 9.042°

b)Active power for path I=𝑉𝐼1𝑐𝑜𝑠Ф1 = 200x1.908 cos(17.30) = 364.3367watt

Reactive power for path I= 𝑉𝐼1𝑠𝑖𝑛Ф1 = 200x1.908sin (17.30) = 113.478VAR

c)Active power for path II=𝑉𝐼2𝑐𝑜𝑠Ф2 = 200x1.975 cos(9.042) = 390.091watt

Reactive power for path II= 𝑉𝐼2𝑠𝑖𝑛Ф2 = 200x1.975sin (9.042) = 62.077VAR

d)I=I1+I2=1.908+1.975=3.883A

Q.No.10. A voltage V=100sin314t is applied to a circuit consisting of 25 Ω resistor

and an 80μF capacitor in series. Determine

a)an expression for the value of the current flowing at any instant

b)the power consumed & the p.d. across the capacitor at the instant when the current

is one-half of its maximum value.

soln given,

R=25Ω C=80 μF Vm=100

∴ Xc =1

ωc=

1

2πfc=

1

314x80x10−6 = 39.8 Ω

Z= √252 + 39.82 = 47 Ω

Im=Vm/Z=100/47=2.13A

Ф =tan – 1(39.8/25)=57°52’ 1.01 radian(lead)

Hence equation for instanteneous current

i= 2.13sin(314t+1.01)

power consumed =I2R=(𝐼𝑚

√2)2𝑥25 = (

2.13

√2)2𝑥25 = 56.7W

The voltage across the capacitor lags the circuit current by 𝜋

2 radians. Hence its equation is

given by Vc=Vcmsin(314t+1.01 – π/2) where Vcm=ImX xC = 2.13x39.8=84.8V

Now, when I is equal to half of maximum current then

i=0.5x2.13A

∴ 0.5x2.13=2.13sin(314t+1.01)

or, 314t+1.01=sin – 1(0.5)=π

6 or 5

π

6radian

∴ vc=84.8sin(𝜋

6−

𝜋

2)=84.8sin(−

𝜋

3)= – 73.5V

or, 84.8sin(5𝜋

6−

𝜋

2)=84.8sin(

𝜋

3)= 73.5V

hence p.d across the capacitor is 73.5V






Q.No.11. It is desired to operate a 100-W,120Velectric lamp at its current rating from

a 240V,50Hz supply. Give a detail of simplest manner in which this could be done

using a)a resistor b) a capacitor c) an indicator having resistance of 10 Ω . what

power factor would be presented to supply in each case and which method is most

economical of power.

soln

Rated current of bulb = 100

120=

5

6A

a) by using resistance

we have p.d across R =240 – 120=120V

∴ R=

120

5

6= 144 Ω

Power factor of the circuit is unity.

Power consumed= 240x 5

6=200W

b)By using capacitor

we have

Vc=√2402 − 1202 = 207.5

Xc= 207.5 x 5

6 =249 Ω

∴ 1

314𝐶= 249 => c=12.8 μF

p.f= cos Ф = 120/240=0.5 (lead)

power consumed=240x 5

6𝑥0.5 = 100W

c) by using an indicator

VR=5

6𝑥10 =

25

3𝑉

∴ VL=√2402 − (120 +25

3

2) = 203V

314L x 5

6= 203

∴ L=0.775

Total resistive drop = 120+25

3= 128.3 V

Cos Ф = 128.3

240= 0.535(lag)

Power consumed = 240 x 5

6 x 0.535 = 107W

As method b) consumes less power , that is most economical.

Q.No.12. A resistance Are, an impedence L =0.01H and capacitance C are connected

in series. When voltage of V=400cos(3000t – 10 ° ) volts is applied to the series

combination, the current is 10√𝟐cos(3000t – 55 ° )A. Find R & C

soln

The phase difference between applied voltage and current circuit is (55° – 10°)=45° with

current lagging. The angular frequency=3000 redian/second. Since current lags XL>Xc.

Net reactance X= XL – XC. also XL= ωL= 3000x0.01=30 Ω

Tan Ф = X

R tan45=

X

R ∴ X=R

Now, Z= VM

IM =

400

10√2 = 28.3 Ω

Z2=R2+X2=2R2






∴ R=𝑍

√2=

28.3

√2= 𝟐𝟎 Ω

X= XL – XC =30 – XC =20

∴ XC =10 Ω 1

𝜔𝐶= 10

∴ C=33μF

Q.No.13. find the values of current I.V1, V2 and p.f of given circuit A and also draw

phaser diagram.

soln

L=0.05+0.1=0.15H

XL=314x0.5=47.1 Ω

XC= 106/314 x 50=63.7 Ω

X=47.1 – 63.7= – 16.6

R=20+10=30 Ω

Z=√302 + (−16.6)2 = 34.3 Ω

I=𝑉

𝑍=

200

34.3= 5.83A

XL1=314x .05 =15.7 Ω

Z1=√102 + 15.72 = 18.6 Ω

V1=IZ1=5.83x18.6=108.4V

Ф1=cos – 1(10/18.6)=57.2° (lead)

XL2=314x0 .1 =31.4 Ω

Xc= – 63.7 Ω

X=31.4 – 63.7= – 32.2 Ω

Z2=√202 + (−32.2)2 = 38 Ω

V1=IZ2=5.83x37.905=221V

Ф2=cos – 1(20/38)=58.2° (lead)

Combined p.f= cos Ф =R/Z =30/34.3 =0.875

Phaser diagram is shown in fig:B






Chapter six Transformer

Transformer & its constuction

A transformer is a state electrical device by means of which electrical energy is

transformed from one circuit to another circuit

without any moving path.

Basically a transformer consists of an iron core

on which two separated cores are wound.

The physical basis of transformer is mutual

induction between two circuit linked by a

common magnetic flux.

It consists of two inductive coils which are

electrically separated but magnetically linked

through the path of low reluctance.

The frist coil in which electric energy is supplied from AC supply is called primary

winding & another from which energy is drawn out is called secondary winding.

The voltage source of one circuit could be higher, lower or equal to voltage level of

other circuit as per need.

Working principle

Consider a transformer whose

secondary wiring is open circulated &

primary wiring is connected to

sinusuidal Alternating voltage V where,

V1=Vmsinωt……………………..(i)

If the coil is purely conductant

then the potential difference between

causes alternating current to flow in primary which lags voltage V1 by 90° i.e I0=

Imsin(ωt-90°)

As secondary winding is open, the primary winding draws a magnetising current

only whose funtion is to magnetise the core. Hence this current will magnetise the core

and It produce a flux Ф which will circulate in the core & will be alternating in nature and

in phase with I0

Ф =Ф msin(ωt-90°)………………………(ii)

As this flux is linked with secondary winding, according to faradays law of

electromagnetic induction emf E2 will be induce in secondary winding. This induced emf

E2 will be antiphase with V1. Also changing flux is linked with primary winding,emf E1

will also induced in primary winding and E1 is in phase with E2. Waveform of voltage,

flux and emf are as follow






EMF equation of Transformer

Suppose,

N1=Number of turns in Primary

N2= Number of turns in secondary

Фm=maxium magnetic flux in core

F=frequency of applied voltage

V2=Terminal voltage across

secondary winding

V1=Primary voltage=Vmsinωt

Then,

Ф =Ф msin(ωt-

90°)…………………(i)

Differentiating equation (i) with respect to time

d∅

dt= ∅mωcos (ωt − 90°)

According to faradays law of electromagnetic induction, the instanteneous value of emf

induced in primary winding is given by

E1 = N1d∅

dt= N1∅mωcos (ωt − 90°)

∴ E1 = N1∅mωsinωt Comparing with V1 =Vmsinωt , we can say that the emf E1 has maxium value at N1∅mω

Hence the RMS value of E1 is given by

E1 =N1∅mω

√2 =

N1∅m2πf

√2= 4.44fN1∅m …………..(a)

Similarly RMS value of e.m.f induced in secondary winding is

E2 =N2∅mω

√2 =

N2∅m2πf

√2= 4.44fN2∅m …………..(b)

Here equation (a)&(b) are e.m.f equation of transformer.

Voltage and current transformation ratio

If we divide equation (b) by equation (i) we get, E2

E1=

4.44fN2∅m

4.44fN1∅m=

N2

N1

∴E2

E1=

N2

N1 =K

neglect the internal resistance of a coil, then

E1≈ V1 E2≈ V2 E2

E1=

V2

V1=

N2

N1= K

Where K is known as Transformation ratio of transformer.

Note

If K>1; N2>N1 , then V2>V1 such transformer is called step up transformer

If K<1; N2<N1 , then V2<V1 such transformer is called step down transformer

If K=1; N2=N1 , then V2=V1 such transformer is called isolated transformer

For ideal loss less transformer

Input VA=Output VA

V1I1=V2I2

∴V2

V1=

I1

I2

∴E2

E1=

V2

V1=

N2

N1=

I1

I2= K






No load operation of transformer

When no load is connected across the secondary winding , I0 current flow through the

primary winding.The ideal transformer cannot be made in real life which donot have any

flux leakage. The coils of real transformer will have some resistance.

The primary input current under no-load condition has to supply

i) Iron loss in the core

ii) A very small amount of cu loss in primary winding.

Hence primary input current I0 is not at 90° nehind V1 nut lags it by angle Ф0 which is

less than 90°.

The phasor diagram for no load operation is shown

above in right side. Here current I0 is resolved into two

components:

Iw=I0cos∅0=component of I0 in phase with V1.This

is known as active or working or iron loss

component.It is responsible for producing heat loss

due to heating of core.

Iμ=I0sin∅0=component of I0 in perpendicular with

V1.This is known as magnetising component.It is

responsible for sustaining alternating flux Ф in the

core.

The resolved component of current I0 has two distinc actions. The effect of this two

component can be represented by circuit model as shown in the figure. No load

equivalent circuit below.

The active power consumed by the transformer at no load is given by:

W0=V1I0cosФ0watts= No load power loss=Iron loss

Where cosФ0= nnoload power factor

It should be noted that no load current I0 is very small and no load copper loss is

negligible

Loaded operation of transformer

When the load is connected across the secondary winding ,secondary current I2

will flow through the load. Hence the secondary winding will produce its own magnetic

flux Ф2 which oppose the main magnetic flux Ф produced by primary winding.

For ideal transformer

Output power=Input power

For loaded real transformer

Output power=V2 I2






Additional current I2′ will flow through winding so that the input power increases from V0

I0 to V1 I1 and there is power balance between primary and secondary winding.At this

stage primary winding produces the additional magnetic flux ∅2′ which is equal and

opposite to Ф2 so that net magnetic flux in the core is again Ф

In actual transformer , the primary and secondary winding will have some

resistance and also they will have some leakage inductance which donot help in the

process of emf inducing but cause reactive voltage drop.Hence equivalent circuit of an

actual transformer can be written as shown in fig below

Here,

V1= supply voltage to primary winding

I0=NO load primary current

Iw=Loss component of I0

Iμ=Magnetising component of Io

R0=Core loss resistance

X0=Magnetising reactant

R1=Resistance of primary winding

X1=Leakage resistance of primary winding

E1=Emf induced in primary winding

E2=Emf induced in secondary winding

R2=Resistance of secondary winding

X2=Leakage reactance of secondary

winding

V2=Terminal voltage across the load

The terminal voltage V2 decreases with the

load current because there is some voltage

drop in R2 & X2 and the terminal voltage is

given by

V2 = E2

− I2 R2 − I2 X2

The phaser diagram of above equation is

shown on right side






Losses in transformer

In a static transformer, there are no friction or windage losses. Hence, the losses

occuring are Iron loss and coppor loss.

Iron loss

It is the summation of eddy current loss and hyteresis loss. The core flux in the

transformer remains practically constant for all loads(its variation being from no load to

full load). The core loss is practically the same at all load.These losses are minimized by

using steel of high silicon content for the core and by using very thin laminations. It is also

known as core loss. Iron loss for a transformer is constant.

Iron loss=Hysteresis loss+eddy current loss

Wi=Wh+We

Hysteresis current loss can be reduced by using high grade silicon steel and eddy current

loss can be reduced by using laminated iron core.

Copper loss

This loss is due to the ohmic resistance of the transformer windings.

cu loss=I2R

Total cu loss=I12R1 + I2

2R2 . Cu loss is proportional to I2.

Copper loss will vary according to load.

Efficiency of a transformer

Efficiency of transformer at a particular load and power factor is defined as the

output divided by input power in watts or kilowatts.

Efficiency =Output power

input power

Efficiency =Output power

Output power + losses=

Output power

Output power + Cu loss + iron loss

or, Efficiency =Input power − losses

Input power= 1 −

Cu loss + iron loss

Input power

It is denoted by ŋ

Condition for maximum efficiency in transformer

cu loss (Wcu)=I2R = I12R1 = I2

2R2 …………………………(i)

Wi=Wh+We

considering primary side

primary input = V1I1 cos ∅1

ŋ =V1I1 cos∅1−Wcu−Wi

V1I1 cos∅1

or,ŋ =V1I1 cos∅1−I21R1−Wi

V1I1 cos∅1

or,ŋ = 1 −I1R1

V1 cos∅1−

Wi

V1I1 cos∅1 …………………………(ii)

Differentiating equation (i) w.r.t I 1

dŋ

dI1= 0 −

R1

V1 cos∅1−

Wi

V1I12 cos∅1

For maximum efficienty R1

V1 cos∅1=

Wi

V1I12 cos∅1

→ Wi = I12R1

∴ Wi = Wcu …………………………(iii)






Similarly

Wi = I22R2

I2 = √Wi

R2

The output current corresponding to maximum efficiency is I2 = √Wi

R2

Hence from above equation (iii) we can conclude that to obtain maximum efficiency in

the transformer iron loss=copper loss

Notes

If we’re given iron loss and full load copper loss then the load at which two losses will

be equal.

Max efficiency load = full load X √Iron loss

Full load copper loss

The efficiency at any load is given by

ŋ =x ∗ full load KVA ∗ power factor

(x ∗ full load KVA ∗ power factor) + Wcu + Wi

Reason for transformer rating in KVA

Copper loss of transformer depends on current and iron loss in voltage. Hence total

transformer loss depends on volt-ampere(VA) and not on phase angle between voltage and

current i.e. it is independent of load power factor. That is why rating of transformer is in

KVA and not in KW.

Solved numerical problems in transformer

Q.No.1. The maximum flux density in the core of the 250/3000V , 50 Hz single phase

transformer is 1.2wb/m2. If the emf induced per turn is 8V, determine

i)the primary and secondary turns ii)Area of core

Solution

Given

E1=250V

E2=3000V

f=50Hz

Βm=1.2wb/m2

Emf induced per turn=8V

Area of core=? Primary and secondary turns=?

We know,

E1=N1 x emf per turn

Or,250=N1 x 8 →N1=250

8

∴ 𝐍𝟏 = 𝟑𝟏. 𝟐𝟓 ≈ 𝟑𝟐

Similarly

E2=N2 x emf per turn

Or,3000=N2 x 8 →N2=3000

8

∴ 𝐍𝟐 = 𝟑𝟕𝟓

We have,E2=4.44fN2Фm → 300=4.44 x 50 x 375 x Βm x A →4.44 x 50 x 375 x 1.2 x A

∴ 𝐀 = 𝟎. 𝟎𝟑𝐦𝟐

Q.No.2. A 25KVA transformer has 500 turns in primary and 50 turns in secondary

winding. Primary winding is connected to 3000V,50HZ supply. Find the full load

primary and secondary current, secondary emf and maximum flux in the core.






Soln,given

P=25KVA=25000VA

N1=500

N2=50

E1=3000V

f=50Hz

we have,

P=E1 x I1 →25000=3000 x I1 → I1=25000

3000 → I1=8.334A

Now, N2

N1=

I1

I2→ I2 =

N1

N2 xI1 =

500

50x 8.334 = 83.334

∴ 𝐈𝟐 = 𝟖𝟑. 𝟑𝟑𝟒𝐀 N2

N1=

E2

E1→ E2 =

N2

N1 x E1 =

50

500 x 3000 = 300V

∴ 𝐄𝟐 = 𝟑𝟎𝟎𝐕

E1=4.44fN1Фm→ Фm=E1

4.44x50xN1 =

3000

4.44x50x500 = 0.027wb=27.02mwb

∴ Ф𝐦 = 𝟐𝟕. 𝟎𝟐𝐦𝐰𝐛

Q.No.3. A 2200/200V transformer draws no load primary current of 0.6A & observes

400W. Find the magnetising & iron loss component of current.

Soln

V1=2200V

V2=200V

I0=0.6A

Power P=400w

We have

P=V1I0cosФ → V1Iw → Iw= 400

2200= 0.182A

∴ 𝐈𝐰 = 𝟎. 𝟏𝟖𝟐𝐀

I0 = √Iw2 + Iμ

2 → 0.62 = 0.1822 + Iμ2 → Iμ = √0.62 − 0.1822 = 0.572A

∴ 𝐈𝛍 = 𝟎. 𝟓𝟕𝟐𝐀

Q.No.4. A 2200/200V transformer takes 0.5A at power of 0.3 on open circuit. Find

magnetising & working component of no load primary current.

Soln, given

V1=2200V

V2=200V

I0= 0.5A

Power factor=cosФ= 0.3

We have

Iw=I0cosФ=0.5 x 0.3

∴ 𝐈𝐰 = 𝟎. 𝟏𝟓𝐀

I0 = √Iw2 + Iμ

2 → 0.52 = 0.152 + Iμ2 → Iμ = √0.52 − 0.152 = 0.476A

∴ 𝐈𝛍 = 𝟎. 𝟒𝟕𝟔𝐀

Q.No.5. In a 25KVA 2000/200V single phase transformer, the iron loss & full load cu

loss are 350 & 400watt respectively. Calculate the efficiency at unity power factor on

a)Full load

b)half load

soln given

We have,

For full load

Iron loss=350watt






Cu loss=400watt

total loss=iron loss+cu loss=350+400=750watt

Full load output @ unity=25KVA x 1=25Kw=25000w

Input =750+25000 =25750W

Efficiency(ŋ) =output

inputx100%

ŋ =25000

25750 x 100%

∴ ŋ = 𝟗𝟕%

For half load

Iron loss=350watt

Half load Cu loss=400 x(1

2)2=100watt

Total loss=350+100=450 watt

Half load output @ unity=12.5 x 1=12.5 Kw= 12500w

Input=450+12500=12950w


inputx100%

ŋ =12500

12950 x 100%

∴ ŋ = 𝟗𝟔. 𝟓𝟐%

Q.No.6. A 5-KVA , 2300/230V, 50Hz transformer was tested for the iron losses with

normal excitation & cu losses at full load and these were found to be 40W and 112W

respectively. Calculate the efficiency Of transformer 0.8 power factor for the following

KVA outputs:

a)1.25

b)2.5

c)3.75

d)5.0

e)6.25

f)7.5

Soln, given

Full load cu loss=112W

Full load iron loss=40W

a)Cu loss @ 1.25 KVA=

112 x (1.25

5)2=7W

Total loss=40+7+47W

Output=1.25 x 0.8=1KW=1000W


output+lossx100%

ŋ =1000

1000+47x100% = 95.51%

∴ ŋ = 95.51%

b)Cu loss @ 2.5 KVA= 112 x (2.5

5)2=28W

Total loss=40+28=68W

Output=2.5 x 0.8=2KW=2000W


output+lossx100%

ŋ =2000

2000+68x100% = 96.71%

∴ ŋ = 96.71%

c)Cu loss @ 3.75 KVA= 112 x (3.75

5)2=63W


Output=3.75 x 0.8=3KW=3000W


output+lossx100%

ŋ =3000

3000+103x100% = 96.68%

∴ ŋ = 96.68%






d)Cu loss @ 5 KVA= 112 x (5

5)2=112W


Output=5 x 0.8=4KW=4000W


output+lossx100%

ŋ =4000

4000+47x100% = 96.34%

∴ ŋ = 96.34%

e)Cu loss @ 6.25 KVA=112x (6.25

5)2=175W


Output=6.25 x 0.8=5KW=5000W


output+lossx100%

ŋ =5000

5000+215x100% = 95.88%

∴ ŋ = 95.88%

f)Cu loss @ 7.5 KVA= 112 x (7.5

5)2=252W


Output=7. 5 x 0.8=6KW=6000W


output+lossx100% =

6000

6000+292x100% = 95.36%

∴ ŋ = 95.36%

Q.No.7. A 200-KVA transformer has an efficiency of 98% at full load. If the maximun

efficiency occurs at three quarters of full load, calculate the efficiency at half load.

Assume negligible magnetizing current and p.f. 0.8 at all loads.

Soln, given

At full load efficiency=98%

Full load output= 200x0.8=160KW

Full load input=160/0.98=163.265KW

Full losses=163.265-160=3.265KW

Let Wcu be cu loss & Wfe be iron loss then

Wcu+ Wfe =3.265KW………………………………….(i)

loss at 75% of Full load= Wcu(3

4)2=

9Wcu

16………...(ii)

As Ŋmax occurs at three quarters of full load when Cu loss becomes equal to iron loss

∴ Wfe =9Wcu

16

Substituting value of Wfe in equation (i), we get

Wcu+ 9Wcu

16 =3.265KW=3265W

∴Wcu =2090W

∴ Wfe=1175w

Half load unity power factor

Cu loss=2090 x (1

2)2= 522W


Output=100x 0.8=80KW=80000


output+lossx100% =

80

80000+1697= 97.9%

∴ ŋ = 97.9%






Chapter seven DC machines

DC Machines

DC machines are rotating electrical machines for continuous energy conservstion

which can be used as motor as well as generator. DC motor converts electrical energy into

mechanical rotation and DC generator converts mechanical energy into electrical energy.

Construction and Parts of DC machines

Field pole & field winding: Field poles are made of laminated silicon steel on

which the emamel insulated copper wire winding are provided.This winding is known as

field winding.

Armature: It is rotating part of machine.This is cylindrical in shape with a central

shaft. The armature is also made of laminated silicon steel sheets.In actual machines an

armature conductor represent many numbers of turns.

Commutator segments: End of armatures conductor are connected to segment

known as commutator segment which are electrically isolated by some insulator like

mica.They are made of copper. The function of the commutator is to facilate collection of

current from the armature conductor

Carbon brushes: The function of these brushes is to collect current from the

commutator & are usually made of carbon graphite. They are in shape of rectangular

block.These carbon blocks are fixed & donot moves while commutator segments rotates

along with the armature.

DC generator:

It is DC machine which converts mechanical energy into electrical energy.

Operating principal of dc generator

The simpler form of DC generator with two

field poles and an armature coil A-B is

given in figure shown beside . When the

armature is continuously rotated the

armature conductirs A and B will cut the

magnetic flux produced by the field poles.

Hence according to Faradays law of

electromagnetic induction, emf will induce

across the coil A-B.The nature of emf

induced will be ac as shown in figure

below






If we could connect a load across coil A-B current will pass through the load. But it is very

difficult to connect the load directly across the rotating armature coil.in order to facilate the

connection of external load across the armature coil, carbon brushes and commutators

segments are used as shown in fig:Dc generator with carbon brush and commutator

segments. Carbon brushes and commutator also helps to convert the ac emf induced in the

armature coil into dc current across the load.

From zero position to 180° rotation, the direction current in the armature conductor ‘A’ is

going inside and the direction through the armature conductor ‘B’ is coming out as

determined by right hand fleming’s rule. Hence C2 collects current from the commulator

segment and delever to load.c1 receive the current back from the load. After 180° rotation

the situation will be as shown in fig @ right side.Here direction of current in the armature

conductors has changed but the C2 is still collecting the current from the armature and

delevering to the load hence the direction of current through the load is unidirectional DC.

Emf equation of dc generator(emf induced in dc generator)

Let us suppose

Ф=magnetic flux per pole

Ρ=number of magnetic poles

Z= Total number of armature conductor

N=Speed of armature in rpm

∴Average emf generator per condcutor=d∅

dt

Magnetic flux cut by each conductor in one revolution= dФ=Фp

Time for one revolution dt =60

Nsec

∴ Average emf generator per conductor=∅PN

60

Let A=number of parallel paths in armature winding

Then number of conductor in series =𝑍

𝐴

Total emf across the brushed are given by

E =∅ZNP

60A






Method of excitation

The field winding needs some dc current to magnetise the field pole. This field current is

known as excitation for the dc generator.The excitation can be provided by various method

and accordingly the dc generator can be classified as follow.

1. Separately excited dc generator

2. Self excited dc generator.

Separately excited dc generator:In

this type of DC generator the field

winding is connected to an external

source of DC supply other than

armature of its own machine.

When the load is connected across

the armature terminals armature

currents Ia will flow which is equal to

the load current IL.The terminal voltage V fall from its open circuit emf E due to a

voltage drop caused by current flowing through armature resistance,Ra. Hence the

terminal voltage on separately excoted DC generator is given by:

V=E-IaRa

If there are voltage drop in brushes then

V=E-IaRa-voltage drop due to brushes

Self excited dc generator: In this type of generator the field winding is excited by

parts of the current generated by armature itself.No external DC source is required for

such generator.The field winding and armature winding have electrical connection self

excited DC generators can be classified into three types which are dscribed as follow.

DC shunt generator:

In DC shunt generator the field winding is

connected in parallel with the armature as shown

in fig:2.This generator shall started without load.

The field winding consisits of residual flux

because of which an emf is generated in

armature. Since no load is connected the total

armature current goes to the field winding and

increases flux which in turn increases the

generation of emf in armature. In doing so, the

production of flux at the field winding reaches to

saturation and load is connected to the generator . Then the field winding draws small

amount of current as it has relatively high resistance than armature resistance.The

current generated by armature Ia divides into two parallel path, one to the field winding

and other to the load. Hence

Ia=If+Ir

Now,

Field current ,If =V

Rf

Load current ,IL =V

RL

Terminal voltage across the load is V=E-IaRa……….(i)






Above equation shows that with increase of current Ia, voltage V across the load

decreases. Fig:3.a shows how the terminal voltage decreases with increasing value of Ia.

this curve is known as load characteristics of DC shunt generator.

The open circuit characteristics of Dc shunt generator is shown in fig:3.b.It shows that the

emf across armature varies with field current.If OA is the magnitude of emf because of

residual flux. Whwn it increases, E also increases from A-B.When flux reaches its

saturation, then E because constant

DC series generator:In DC series generator the field winding is connected in series

with the armature winding as shown in fig:4.This type of generator shall be started

with the load. These generators are rarely used in practise. Here the figure shows that

the same current flows through

armatures, field and the load. The

terminal voltage ‘V’ across the load is

given by :

V=E-IRf – IRL

The load characteristic of series generator is shown in fig:5. When the current load

increases the emf will also increase. Hence series generator has rising voltage

characteristic but at overload condition the voltage starts decreasing.

DC compound generator: This type of DC generator have two sets of field winding

one of them is connected in series with armature winding and other is connected in

parallel with armature winding.The series field winding is made from thick wire with

few turns and shunt field winding is made from thin wire with many numbers of

turns.There are two types of DC compound generator as shown below






Long shunt generator: In long shunt genrator the shunt field winding is

connected across armature together with series field winding. For long shunt

generator

𝐈𝐟 =𝐕

𝐑𝐬𝐡 𝐈𝐚 = 𝐈𝐟 + 𝐈𝐋

V=E-IaRsh-IaRse

Short shunt generator:In where as in the short shunt generator the shunt field

winding is connected across the armature winding only.

𝐈𝐟 =𝐕𝐬𝐡

𝐑𝐬𝐡=

𝑬−𝑰𝒂𝑹𝒂

𝐑𝐬𝐡 𝐈𝐚 = 𝐈𝐟 + 𝐈𝐋

V=E-IaRa-ILRse

The characteristic of compound generator lies between the characteristics of shunt

and series generators. Fig:7 shows the load characteristics of compound generator.

Armature reaction:Since a current carrying conductor produces magnetic field around

it,the armature conductor will also produce its magnetic field around it.The effect of

magnetic field produced by armature conductor on magnetic field poles is known as

Armature reaction.






DC motor:It is a dc machine that converts electrical energy into mechanical rotation.The

electrical energy is given across the armature and field windings and the armature

produces the mechanical rotation.

Operating principle of DC motor

In DC motor the electrical energy is supplied across field winding and armature

winding’s as well. When field winding is supplied by DC voltage, the field current will

flow through the field poles winding this current will magnitize the field poles resulting a

magnetic field in the space betweeen two poles.

When the armature conductors are also supplied by DC current through carbon

brushes. These armature conductors interact with magnetic field produced by the field

poles and force will develop on the armature conductors. This force will produce

continuous rotation of armature.

Back emf in DC motor

When the armature of dc motor rotates the

armature conductor cuts the magnetic flux

produced by the field poles. Hence according to

faradays law of electromagnetic induction emf

will induced acros the armature conductors. The

direction of this emf is opposite to the applied

voltage ‘V’. This emf is known as back emf.

The field current is given by If =V

Rf

Using kifchhoff’s law we can write

V − IaRa − Eb = 0

∴ Eb = V − IaRa

The magnitude of back emf in DC motor is given by

Eb =∅ZNP

60A

The magnetude of back emf is always less than the applied voltage.

Role of back emf

Back emf protects the armature from short circuit during rumming condition

Back emf helps tge motor to prodce required amount of torque qccording to increased

or decreased external load torque (i.e it acts as current ontroling agent)

Back emf acts as energy converting agent. The back emf in dc motor acts as the

opposing agent due to which the Dc motor is able to convert electrical energy into

mechanical rotation.

Types of DC motor

DC shunt motor:In Dc shunt motor the field

wiring is in parallel with the armature across the

supply as shown in figure.

The armature current is flowing into the armature

against the opposition of back emf Eb and

armature develops the mechanical power to EbIa

watt.






Let Ta=torque developed by the armature

Then, 2πNTa

60= EbIa

But Eb =∅ZNP

60A

or2πNTa

60=

∅ZNP

60A Ia

or, Ta =ZP∅Ia

2πA

∴ Ta ∝ ∅Ia

Using kifchhoff’s law we can write

V=Eb+IaRa

Ia =v−Eb

Ra

Armature torque- armature current characteristic

Since field current is constant in dc shunt motor, the magnetic flux is also constant.

Therefore Ta ∝ Ia .

Here torque increases proportionally with the armature current

Speed-torque characteristic

When external load on shaft of dc motor increases the speed will decrease. Them

back emf will be decrease resulting more armature current. Hence more torque will be

developed to overcome the increase load on shaft

DC series motor: In Dc series motor the field wiring is in series with the armature

across the supply as shown in figure.

In this motor current is flowing through both the armature and field winding.

If = Ia

𝑜𝑟, Ta ∝ ∅Ia

∴ Ta ∝ Ia2

Hence armature torque is propertional to the square of armature current.


The torque is directly proportionel to current over limited range before magnetic

saturation is reached.

Thus

Ta ∝ ∅Ia ∴ Ta ∝ Ia2

After magnetic saturation Ta ∝ Ia







At starting the speed is very low and back emf is also very small .Therefore the

motor draws very high current which produced strong magnetic flux. Since Ta ∝∅Ia the motor produce high torque at lower speed.

DC compound motor: These motor have both series and shunt field winding. There

are two types of dc compound motor. They are as follow:

Commulative compound:In this type of compound motor the series winding

produces the magnetic flux in same direction as produced by the shunt field

winding .

Differential compound: In this type of compound motor the series winding

produces the magnetic flux in the opposite direction to that produced by shunt

field winding.

Characteristic of compound generator



Solved numerical examples

Q.No.1. An 8 pole wave connected armature has 600 conductor and is driver at 625

rpm. If the flux per pole is 20 μwb. Determine the generated emf.

Given,

P=8

Z=600

A=2

N=625

Ф=20μwb=20 x 10-3wb

E=?

We know that,

E =∅ZNP

60A=

20 x 10−3x 600x625x8

60x2= 500𝑣𝑜𝑙𝑡𝑠






Q.No.2. A 4 pole generator has a lap wound armature with 50 slats with 16 conductor

per slot. The useful flux per pole is 30 μwb. Determine the speed at which the

machine must be driven to generate an emf of 240V.

Given,

P=4

A=4

Ф=30 μwb=30x10-3

Z=50 x 16=800

E=240V

N=?

We know that,

𝑁 =𝐸𝐴

∅𝑍𝑃=

240 𝑥4

30 𝑥10−3𝑥800𝑥4= 10𝑟𝑝𝑚

Q.No.3.Determine the terminal voltage of a separately excited generator which

develops an emf of 200V, and has an armature current of 30A on load. Assume that

armature resistance is 0.30Ω.

Soln,given

Ia=80A

E=200V

IL=30A

Ra=0.30Ω

V=?

We have

V=E-IaRa=200 -30x 0.30= 200 – 9.0 =191V

∴ V = 191V

Q.No.4. A separately excited generator is connected to a 60Ω load and a current of

8A flows. If the armature resistance is 1 Ω. Determine

a)terminal voltage b) generator emf

soln, given

Ia=IL=8A

RL=60Ω

Ra=1Ω

V=?

E=?

We have

V=ILRL=8x60

∴ V = 480V

Again

E=V+IaRa=480+(8x1)

∴ E = 488V

Q.No.5. A dc shunt generator delivers 50A to the load at 220 volts. Calculate the emf

generator by the armature. Given that the armature winding resistance is 110 Ω.

Soln, given

IL=50A

Ra=0.8Ω

Rf=110Ω

E=?

Now,

If =v

Rf=

220

110= 2A

𝐼𝑎 = 𝐼𝑓 + 𝐼𝐿 = 2 + 50 = 52𝐴

𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 = 220 + 52𝑥0.08 = 220 +4.16

∴ E = 224.16V






Q.No.6. A long shunt compound generator delivers a load current of 40A at 400V

and has a armature, series field and shunt field resistances of 0.07Ω,0.04 Ω & 200 Ω

respectively. Calculate the generated voltage and armature current.

Soln, given

V=400V

IL=40A

Rse=0.04 Ω

Rsh=200 Ω

Ra=0.07 Ω

Ia=?

E=?

We have

If =V

Rsh=

400

200= 2A

Ia = If + IL = 2 + 40 = 42𝐴

𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 + IaRse = 400 + 42𝑥0.07 + 42𝑥0.04

∴ E = 404.62𝑉

Q.No.7. A short shunt compound generator delivers a load current of 28A at 220V

and has armature , series and shunt field resistance of 0.06Ω,0.25Ω & 240Ω

respectively. Calculate the induced emf and armature current.

S0ln,given

V=240V

IL=28A

Rse=0.25 Ω

Rsh=240 Ω

Ra=0.06 Ω

Ia=?

E=?

We have

Vsh = V + ILRse = 220 + 28x0.25

∴ Vsh = 227V

If =Vsh

Rsh=

227

240

∴ If = 0.95

𝐼𝑎 = 𝐼𝑓 + 𝐼𝐿 = 0.95 + 28 = 28.95𝐴

𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 + IaRse

= 220 + 28.95𝑥0.06 + 28𝑥0.25

∴ E = 228.73𝑉

Q.No.8.A short shunt cumulative compound dc generator supplies 7.5Kwat 230V.

the shunt field , seris field and armature resistances are 100 Ω,0.3 Ω and 0.4 Ω

respactively. Calculate the induced emf and the load resistance.

soln, given

p=7.5Kw=7500w

Rsh=100Ω

Ra=0.4 Ω

Rse=0.3 Ω

V=230V

E=?

RL=?

Now,

P=VxIL

0r,7500=230x IL

∴ 𝐼𝐿 =7500

230= 32.61Ω

We know,






V=IL xRL

Or,230=32.61 x RL

∴ 𝑅𝐿 =230

32.61= 7.053Ω

Vsh = V + ILRse = 230 + 32.61x0.3

∴ Vsh = 239.78V

If =Vsh

Rsh=

239.78

230

∴ If = 2.39

𝐼𝑎 = 𝐼𝑓 + 𝐼𝐿 = 2.39 + 32.61 = 35𝐴

𝐸 = 𝑉 + 𝐼𝑎𝑅𝑎 + IaRse

= 230 + 35𝑥0.07 + 32.61𝑥0.3

∴ E = 253.78𝑉

Q.No.9. A dc motor operates from a 240V supply, the armature resistance is 0.2Ω.

determine the back emf when the armature current is 50A.

Given,

V=240V

Ra=0.2 Ω

Ia=50A

Eb=?

We know

Eb=V-Ia Ra=240 – 50x 0.2=240 – 10

∴ Eb = 230V

Q.No.10. A dc machine has an armature resistance of 0.5 Ω. If the full load armature

current is 20A. Find the induced emf when machine acts as

1)generator 2)motor

Given

V=220V

Ra=0.5 Ω

Ia=20A

Eb=?

We have

1)when machine acts as generator

Eb = V + IaRa = 220 + 20x0.5

∴ Eb = 230V

2)when machine acts as motor

Eb = V − IaRa = 220 − 20x0.5

∴ Eb = 210V

Q.No.11. A 240V shunt motor takes a total current of 30A.If the field winding

resistance Rf =150 Ω and armature resistance 0.4V. Determine the back emf and

current in armature.

Given

V=240V

I=30A

Ra=0.4 Ω

Rf=150 Ω

Eb=?

Ia=?

We know

𝐼𝑓 =𝑉

𝑅𝑓=

240

150= 1.6𝐴

𝐼𝑎 = I − If = 30 − 1.6 = 28.4𝐴

𝐸 = 𝑉 − 𝐼𝑎𝑅𝑎 = 240 − 28.4𝑥0.4 = 240 − 11.36

∴ E = 228.64V






Chapter Eight AC machines

Induction motor

It is an electrical motor which operates from AC voltage source and operates under

the principle of electromagnetic induction. It is also known as asynchronous motor.

Construction of Induction motor

An induction motor consists of three parts: a)Strator b)Rotor c)yoke

Stator: It is stationary part of motot that is made up of circular stamping.Stator winding

are placed on the inner circumference of strator core.Generally three phase winding are

provided on these slots and each phase windings are spaced 120° electrically apart.

Rotar: It is the central rotating part of induction motor which is cylindrical in space with a

central shaft. The shaft is supported by bearing at the both ends so that the rotor rotates

freely. It is made up of laminated silicon steel. There are two types of rotor.

1. Squiral cage rotor: It is made up of cylindrical laminated core with parallel slots

nearby the outer circumference. These parallel slots carry rotor conductors and the end

of these conductors are short circulated by coppering known as ring.

2. Phase wound rotor: It is also made up of laminated core but it has open slots along

the circumference on which three phase winding are provided with same number of

pole as that in the stator winding. The three ends of rotor winding are connected to the

three separate slip rings and the slip rings are short circuted by the carbon brushes with

or without external resistance.

Yoke:It is the outermost frame of the machine. It houses the stator core and provides

mechanical protection for the whole machine.the outer surface of the yoke have many

number of fins to cool the machine.

Operating principal of induction motor

When the three phase stator are supplied by the three phase voltage source, three

phase current will flow through stator windings. These three phase current will magnetize

the strator core. Each winding will produce their own magnetic flux which are 120° apart.

The net flux of the induction motor will be equal to 1.5Фm

Where, 1.5Фm = maximum flux. The stator winding produces rotating magnetic field and

the speed of the rotating magnetic field is given by,

Ns =120f

P

Where, f=frequency of voltage applied across the stator winding

P=number of magnetic poles for which stator winding is wound

This speed is known as synchronous speed(Ns)

The rotating magnetic field produced by the stator cuts the rotor conductor and

hence emf will induce on the rotor conductor according to faradays law of electromagnetic

induction. As the rotor conductor are short circulated current will circulate within the rotor

conductor. Now the current carrying conductor are lying in the magnetic field produced by

strator. Hence force will developed on the rotor conductorand therefor the rotor starts

rotating under the action of this force

The rotor will try to catch up the speed of rotating magnetic field(Ns) but it never

success to do so and always runs at speed less than the synchrinous speed(Ns). Therefor an

induction machine is also called asynchronous machine.






Slip

The fraction by which the speed of the rotor is less than the synchronous speed is

known as slip(s). it is given by

S =Ns−N

Ns

Slip is usually expressed in percentage.

Synchronous speed Ns =120f

p

Synchronous Generator

It is an ac rotating machine which has to be driven at constant speed equal to

synchronous speed. They are also called alternators as it produce alternate voltage.

Construction of synchronous machine(both generator and motor)

The main parts of synchronous machine are described as follow:

i) Strator: It is stationary part of motot that is made up of circular stamping.Stator

winding are placed on the inner circumference of strator core.Generally three phase

winding are provided on these slots and each phase windings are spaced 120° electrically

apart.

ii) Rotor: It is rotating part of machine with number of magnetic poles excited by dc

source from excitor. There are two types of rotor.

a. Salient pole rotor: Its construction is easier and cheaper than cylindrical rotor. It

is mainly used in generator driven by low speed prime movers such as water

turbine, diesel engine etc

b. Cylindrical type rotor: It has smooth magnetic poles in form of a closed cylinder.

Its construction is more compact and robust with compare to salient pole rotor.

They are generally used in generator driven by high speed prime movers like steam

turbine, gas turbine etc

iii) Exciter :Exciter is a felf exciteed dc generator mounted on the shaft of the alternator. It

is supplied dc current to field winding of rotor.

Operating principal of synchronous generator

In synchronous geberator the field poles are rotating and armature conductor are

stationary. The shaft of the machine is driven by prime mover at a constant speed equal to

synchronous speed. The exciter built uo its voltage byself excitation and supply dc current

to the field winding of alternator. The magnetic flux produced bythe rotor poles will cut

the stationary three phase stator winding. Hence according to faradays law of electro

magnetic induction, three phase emf will be induced in stator winding.

Advantages of rotating magnetic system and stationary armature system

The magnetic field system in synchronous generator is opposite to that in dc

generator. Following are the advantages of rotating magnetic system and stationary

system:

The output current can be led to the load directly from the fixed terminals on the stator

without slip rings and brushes.

It is easier to insulate stationary armature winding for high voltage (usually 11kv or

higher) rather than rotating armature






The field winding deals with low current at low voltage. Therefor the rotating field

winding can be easily insulated. Also slip ring and brushes do not have to handle large

current so that the aparking problem at the slip rings minimum.

Parallel operation and syncronization

In power system , two or more alternators in parallel is known as synchronization.

The numbers of alternators are connected through bus bar(usually infinite bus bar). An

infinite bus bar is the bus bar whose voltage and frequency is independent and constant

with load. The following conditions have to be satisfied for synchronizing an alternator

The terminal voltage of both alternator should be equal.

The frequency of both alternator must be equal

The waveform of emf generated by both alternators should be in phase

The percentage impedance of both alternators should be same.

The phase sequence of both alternators must be same

When two alternators are operating ,so that all above requirements are fulfilled.

They are said to be in synchronism. The proces of connecting them in synchronism called

as synchronizaton.

Synchronous motor

It is an ac motor which always which rotates at constant speed equal to

synchronous speed. Some characteristic features of synchronous motor are as follow.

It ran either at synchronous speed or not at all. i.e. while running it maintains a

constant speed equal to the synchronous speed.

As it is not self starting. Some auxillary means has to be used to start thr motor.

The motor can be operated at wide range of power factors both lagging and leading

Operating principle

When the stator winding are supplied by three phase voltage , rotating magnetic

field will produce. At starting the rotor of field winding are unexcited and the rotor is

driven at synchronous speed by some auxilliary means. Then the rotor field winding are

supplied by dc current and the auxilliary means axis disconnected. The rotor pole and

stator pole will get engaged with a strong force and the rotor continuously rotates with

synchronous speed.

Solved numerical problems

Q.No.1.A 6 pole, 50Hz squirrel cage induction motor runs at the speed of 970rpm.

Calculate

i)Synchronous speed

ii)slip

iii)the frequency of rotor current

Soln

Given

P=6

f=50Hz

N=970rpm

Ns=?

S=?

f’=?

now,

Ns =120xf

p=

120x50

6

Ns = 1000rpm






s =Ns−N

Ns=

1000−970

1000= 0.03

𝑠 = 3%

f’ = sf = 0.03x50

f’ = 1.5Hz

Q.No.2. A 6 pole 3-Ф 60Hz induction motor runs at 4% slip at certain load.

Determine

i)synchronous speed

ii)rotor speed

iii) frequency of rotor current

soln

p=6

f=60Hz

s=4%=0.04

Ns=?

N=?

f’=?

now,

Ns =120xf

p=

120x60

6

Ns = 1200rpm

𝑠 =𝑁𝑠−𝑁

𝑁𝑠=

1200−𝑁

1200= 0.04 ↔ 1200 − 𝑁 = 48

𝑁 = 1152𝑟𝑝𝑚

f’ = sf = 0.04x60

f’ = 2.4Hz

Q.No.3. A 4 pole 3-Ф 50Hz induction motor runs at 1440 rpm. Determine the

percentage slip of induction motor.

Soln

p=4

f=50Hz

N=1440rpm

S=?

Now,

𝑁𝑠 =120𝑓

𝑝=

120𝑥50

4= 1500𝑟𝑝𝑚

𝑠 =𝑁𝑠−𝑁

𝑁𝑠=

1500−1440

1500=

60

1500= 0.04

𝑠 = 4%




Documents

Chapter one Introduction Energy sources The various forms