01 35 Current Electricity Study

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SrivastavasSrivastavas St udy Pac k age

CURRENT ELECTRICITY

Electrical Conduction :Is completely analogous to the thermal conduction. As the heat energy flows from higher

temperature to the lower temperature the charge also flows from higher potential to the

lower potential. In fluids (liquids and gasses) ions participate in electrical conduction. In

solid conductors free electrons are responsible for the flow of current in the conductor.

Free electrons are the electrons, which do not remain associated with their parent nuclei

inside the conductor and can move freely inside the material boundary.

Electric Current :

The rate of flow of charge through any cross-section of the conductor is defined as

electric current. It is a scalar quantity.

dtdqi ,

Unit of electric current is Cs1 (Coulomb per second) or A (ampere). It has been accepted

as one of the fundamental units.

The current density J is defined as the current flowing per unit area of cross-section of

the conductor or in other words, current iis the flux of the current density.

sJ.diConventionally the direction of current is in the direction of flow of positive charge or

opposite to the direction of flow of negative charge (electrons).

Drift speed (vd) of free electrons :The free electrons in the conductor keep moving randomly with high speeds, but due to

this motion there is no net transfer of charge through any cross-section. These free

electrons keep colliding with some other electrons or nuclei in the process of motion and

there direction of motion keep changing randomly.

m

eE

drift

E

Relaxation time :

The average time interval between two successive collisions is called relaxation time .

Mean free path :The average distance traveled by the free electrons between the two successive collisions.

dv .

When the potential difference is applied across the conductor the electrons feelaccelerated opposite to the field. Due to this they get displaced slightly opposite to thefield in every free path motion. This way the electrons keep drifting. The random motion

speed of the electrons is 106 m/s, where as the drift speed is 102 m/s. This drift

motion can be assumed as superimposed on the random motion of the electrons. We canalso safely assume that in every collision the kinetic energy due to drift speed is


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Current Electricitycompletely lost by the electrons (the drift speed becomes zero after every collision). Let

there be n free electrons per unit volume of the conductor. consider the length

numerically equal to vd, of the conductor. If the electrons are moving with speed vd, all

the free electrons in the volume Avd will cross the section A in one second. Thus thecurrent in the conductor is charge crossing the section in one second,

vd

A+

dd neAvneAvi

neA

ivd

Ohm's Law :States that the current density in a conductor is directly proportional to electric field

across the conductor. EJ

EJ

,

where is known as the specific conductance or conductivity of the material ofconductor.

EJ

l

V

A

i

iRA

li

A

liV

iV It is the direct consequence of Ohms law. The devices, which follow this rule are calledOhmic devices.

Here

1

is known as the specific resistance or resistivity of the material of

conductor and

A

l

A

lR

is the resistance of the given conductor.

[R] = [ML2T3A2]

Reciprocal of resistance is called conductance G.

VI

RG 1 and denoted by also called mho (ohm1) or siemen (S).

Verification of ohm's law :

A potential difference ofVvolts is applied across the conductor, then the field inside theconductor is

V+

l

SCO 18, 2nd Floor, Sec . 20 D, CHANDIGARH.

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Current Electricity


l

VE

thus the acceleration of free electrons is

ml

eV

m

eEa

In the time between two successive collisions, the speed gained by the electron is

tml

eVv 0

but vd is the average speed gained by the electrons, thus

n

t

ml

eV

n

vvd

ml

eVv

d

!

where is the relaxation time.

SinceneA

ivd therefore

mleV

neAi

A

l

ne

miV

2

m

ne

l

V

A

i 2

m

neEJ

2

Here mand e, the mass and charge of electron are universal constants, n, the number of

free electrons per unit volume depends upon the material of the conductor, and therelaxation time is also a constant which depends upon the material of the conductor aswell as the temperature of the conductor. Thus the quantity

m

ne2

is known as the specific conductance or the conductivity () of the medium. ThereforeEJ

Effect of temperature on conductivity :

! another approach presented by some Indian authors for the calculation of drift speed of

electrons is as follows.In the time between two successive collisions, the displacement of electrons is

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2

2

10 as

hence the average speed is

2

asv

ml

eVv

d2

but in our opinion this is wrong. The drift velocity of electrons is the 'number average'of velocitycomponent of free electrons due to electric field, for large number of electrons at a particular

instant. For this ''Drift speed and collision time'' by Donald E. Tilley, American Journal ofPhysics, June 1976, page 597 can be referred.


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Current ElectricityIn case of metals, with increase in temperature n (the free electron density) does not

change appreciably, therefore when temperature increases due to increased thermal

agitation the possibility of collision increases hence the average time interval between

two successive collision () decreases, thus

2ne

m

m

ne

2

and

Ane

mlR

2 increase or decreases.

for conductors

for semiconductors

In case of semiconductors, initially the increase in temperature increases the free

electron density as the electrons jump from valance band to conduction band, whichover comes the effect of decrease of relaxation time, hence the resistivity decreases

(conductivity decreases), later when a certain temperature is reached the nalmost stops

increasing but still keeps decreasing hence starts increasing (starts decreasing).

Mobility of charge carriers :Mobility of charge carriers is defined as the drift speed per unit strength of electric field

in the conductor

E

vd

j

neA

i

as from Ohm's law Ej , therefore

ne

ne

Temperature coefficient of resistivity & Resistance :Is defined as the change in resistivity per unit resistivity per unit change in temperature.

d

d


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Current Electricity


10

Similarly R the temperature coefficient of resistance is defined as

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dR

dRR

RRR 10

Relation between

and R :

The relation between resistance of a conductor and its resistivity is

A

lR

taking log and differentiating

)ln()ln()ln()ln( AlR

dA

dA

dl

dl

d

d

dR

dR

R

Here is the coefficient of linear expansion of material of the conductor and the

coefficient of superficial expansion (= 2)

2R

R

Equivalent temperature coefficient of resistance :

Series :

Let the resistance R1, R2, ..... , Rn are connected in series with respective thermal

coefficients 1, 2, ..... , n then

RReq

dt

dR

dt

dReq

Rdt

dRR

dtR

dRR

eq

eq

eq

RR eqeqParallel :

Let the resistances R1, R2, ...., Rn are connected in parallel with respective thermal

coefficients 1, 2, ....., n then

RReq

11

dtR

dR

dtR

dR

eq

eq

22

RReq

eq

Effects of Current :There are three effects of current.

a) Thermal effectb) Chemical effect


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Current Electricityc) Magnetic effect

a) Thermal effect :

The work done in moving a charge dqby a potential difference ofVisdqVdw

dt

dqVP

dt

dw

R

VRiViP

22 and the heat generated

While comparing power dissipated in various elements connected in series the current

through elements remains same hence is used. If two or more elements are

connected separately to the same supply or the elements are connected in parallel the

potential difference across them remains same hence

RiP 2

RVP is used to calculate the

power dissipated through them.

2

.

iVdtH

Two elements when connected individually across a supply of constant potential Vthey

produce powers P1 and P2 respectively.2

1

1

VP

R and

2

2

2

VP

R

2

1

1

VR

P and

2

2

2

VR

P

Power produced when their series combination is connected to same power supply

2

1 2

VPR R

2

2 2

1 2

VP

V VP P

1 2

1 2

P PP

P P

Power produced when their parallel combination is connected to same power supply2V

P

R

2

1 2

1 1P V

R R

1 2P P P

Representative example 1 :

Two heating coils can boil separately certain amount of water in the time t1, and t2

respectively when they are separately connected to the domestic supply. What time will they

take to boil same amount of water when they are used simultaneously in

a) series,

b) parallel with the same domestic supply.


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Current Electricity


Solution :

Let R1 and R2 be the resistances of the two heating coils then if the amount of heat

required to boil the water is Qthen

1

1

2

tR

VQ and

2

2

2

tR

VQ

Now when the two coils are connected in series across the same domestic supply then

stRR

VQ

21

2

stt

QVt

QV

VQ

2

2

1

2

2

21 ttts

Similarly when the two coils are connected in parallel to the same domestic supply then

p

eq

tR

VQ

2

ptRR

VQ

21

2 11

ptt

Q

t

QQ

21

21

21

tt

tttp

Maximum Power Theorem :In an electrical circuit, the maximum power can be drawn from the battery when

external resistance is same as the internal resistance of the battery.

Current through the circuit shown is

rE

i

R

rR

Ei

hence, power drawn in the external resistor is

2

22

)( rR

RERiP

0

24

2

2

rR

rRRrRE

dR

dP

rR Therefore maximum power delivered to the external resistance is

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MARATHON-1010

Current Electricity r

E

r

rEP

42

2

2

2

max

In this situation the battery works at 50% efficiency.To increase the efficiency of the battery, power developed through the internal resistanceof battery should be reduced. For this current drawn from battery must be reduced. But

at the same time this will also reduce the power in external resistance. Hence for batteryto operate with better efficiency external resistance should be quite large compared tointernal resistance.

Fuse wire :Is a wire element made of a material with low resistivity and low melting point (mostlymade of an alloy of tin and led) with suitable radius of cross section, connected in the

series of circuit. The fuse wire blows out when current exceeding the limit to which thewire is made protecting the electrical appliances against the current surge.The rate at which heat is produced in the fuse wire must be same as the rate at which itis lost in the form of radiation (considering fuse wire radiating as perfect black body). If

melting point of the fuse is Tand the maximum current rating of it is ithen

2

2242r

liRiTrl

432

2 Tr

i

clearly maximum current rating (i) of the fuse wire does not depend on the length of the

wire and

23

ri ,2Ti

b) Chemical effect :

The process of electrolysis and chemical electrode potential are the chemical effects ofcurrent.

Farade's laws of electrolysis :

i) In the process of electrolysis the mass of ions released on any electrode of thevoltammeter is directly proportional to the amount of charge passed through thevoltammeter.

qm

whereZqm

q

m

Z Z is the electrochemical equivalent of the substance, which is equal to the mass of ionsreleased when one coulomb charge is passed through the voltammeter.

ii) In the process of electrolysis if same amount of charge is passed through severalvoltammeters then the mass of ions released on any electrode of the voltammeter isdirectly proportional to chemical equivalent weight of the substance.

Em EZq

EZ

FZ

E

,


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Current Electricity


Fis the universal constant known as the Farade's number, which is equal to one mole ofcharge.

.NeF The Farade number is the amount of charge needed to liberate one equivalent weight ofthe substance at any electrode of a voltmeter.Since one equivalent weight of the substance will contain

number)s(Avogadro'(valency)

1 NV

n

molecules thus the amount of charge needed to liberate these ions is.FNenVeq

c) Magnetic effect : (This will be discussed in electromagnetism)

THERMO ELECTRICITY :When electrical energy can be converted in to thermal energy, principal of physicalsymmetry says that thermal energy can also be converted in to electrical energy.

Thermocouple : The arrangement of two metallic wires connected at two ends forming two junction(nodes) is called thermocouple.

A

Thermoelectric effect (Seebeck effect) :The phenomenon of production of an electric current in a thermo couple by keeping itstwo junctions at different temperatures is called Seebeck effect (thermoelectric effect).

The magnitude and direction of thermo emf depends on the metals and the temperaturedifference between the two junctions.

Thermoelectric series :Metals placed in order as antimony, nichrome, iron, zinc copper, gold, silver, lead,aluminium, mercury, platinum, nickel, constantan and bismuth form a thermoelectricseries. At the cold junction the current flows from the metal earlier in the series to themetal later in the series.

Contact potential :

When two metal wires are connected at a point, due to net diffusion of free electrons

from the metal of higher free electron density to the metal with lower free electrondensity the metal with higher free electron density becomes a little positive and the othermetal becomes a little negative as it receives the electrons due to thermal diffusion.Potential difference hence developed between the metals connected at a point is calledcontact potential. The difference in rate of diffusion due to difference in temperature atthe two junctions of a metal pair is the cause of Seebeck effect.

Variation of Seebeck emf with temperature :

If the temperature of cold junction is oC and of hot junction is Tthen, the variation of

thermo emf is found to obeycT

22

1 TTE

Where

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MARATHON-1012

Current ElectricitycTTT

Tn

TiThot

Tc

Thermoelectric Power : The rate of change of thermo emf with respect to temperature is defined as thermoelectric power also known as Seebeck coefficient.

TdT

dES

Here and are the constants for a given pair of metals forming a thermocouple.

i) When the two junctions are at the same temperature the thermo emf is zero.

ii) The thermo emf increases with increase in temperature of hot junction and becomes

maximum at a particular temperature called neutral temperature Tn.

The temperature of hot junction at which the thermo emf in a thermocouple is maximum is

called neutral temperature of the thermocouple. At neutral temperature the thermoelectricpower is zero. The neutral temperature is

(a) independent of the temperature of the cold junction.(b) constant for pair of metals forming thermocouple. For Cu-Fe it is 275 oC.

iii) After achieving the maximum value thermo emf starts decreasing with increase in

temperature of hot junction and becomes zero at a temperature Ti called temperatureof inversion.

The temperature of hot junction at which the thermo emf becomes zero again and changes

direction just beyond is called temperature of inversion. The temperature of inversiondepends upon

(a) the temperature of the cold junction.(b) the nature of the metals forming thermocouple.

Relation between Tnand Ti :Thermo emf varies with temperature according to

22

1 TTE

From the definition of neutral temperature

0dT

dES at nTT , hence

0 cn TT

cn TT (1)

Thermo emf becomes zero at the temperature of inversion, hence

02

1 2 cici TTTT

or0 ci TT 2 ci TT


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Current Electricity


at the temperature thereforeci TT

2 ci TT (2)

from (1) and (2)

cnci TTTT 2

2cin

TTT

(3)

Peltier effect :

It is the converse of Seebeck effect. It states that if a current is passed through a junction of two different metals heat is either evolved of absorbed at that junctiondepending on the direction of current.If the direction of current in a thermocouple is reversed the junction at which the heat

was being evolved starts absorbing heat and at the other junction where heat was beingabsorbed starts evolving heat, hence the Pertier effect is reversible.

coldhot

i

Seebeck effect

heatabsorbed

i

Peltier effect

heatevolved

A

A

Peltier coefficient () : The quantity of heat absorbed or evolved at the junction of two different metals whenunit amount of charge is passed through the junction is called Peltier coefficient.

dq

dH J C1

Peltier coefficient depends uponi) nature of metals forming the junctionii) temperature of the junction. Therefore Peltier coefficient is different at the two

junctions of a thermocouple at different temperatures.

Thomson effect :When different parts of a single conductor are maintained at different temperatures due

to temperature gradient heat is evolved or absorbed along the length of the conductor.This heat is over and above joules heating.When heat is absorbed for current flowing in the direction of positive gradient oftemperature (increasing temperature) and evolved for current flowing in the direction ofnegative gradient of temperature (decreasing temperature) the Thomsons effect is said tobe positive otherwise negative.

Thomson coefficient () :It is the amount of heat energy evolved or absorbed when one coulomb of charge ispassed through a conductor whose ends are maintained at a temperature differenceof 1 K.

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MARATHON-1014

Current Electricitydq

dH

T

1

It is also defined as the emf developed between the ends of a uniform conductor whenends of the conductor are maintained at a temperature difference of 1 K.

dT

dV

Free electron theory could not explain zero Thomson effect for Pb and negative Thomsoneffect for Fe, Bi, Co, Pt etc.

Relation between thermo electric coefficients (S, and ) :Following relations between thermoelectric coefficients can be easily established.

TS (4)

dT

dST (5)

Thermoelectric laws :

There are two laws which are experimentally established.

a) Law of successive metals :If a number of different metals form a chain, then the effective emf between extrememetals when placed in direct contact at a given temperature is the sum of the

individual emf between adjoining metals provided all the junctions are at the sametemperature.

Z

Y

D

C

C

B

B

A

Z

A EEEEE ...

b) Law of successive temperatures :

The effective emf of a given thermocouple for the temperatures of junctions as T1 and

Tn is equal to the sum of the emf of same thermocouple for temperature of the

junctions as (T1, T2), (T2, T3), (T3, T4) (T(n1), Tn).n

n

n T

T

T

T

T

T

T

T

T

T EEEEE 14

3

3

2

2

11...

Resistance combination :In a circuit four kinds of connections are possible. A series connection, parallelconnection, star connection and a delta connection.

Series :

When two or more elements are so connected that no branch is coming out of thejunction between the two elements is called a series connection. The ascension condition

for the series connection is that the elements should bear the same current.

E

i

VnV2V1

RnR2R1

nVVVE .....21

neq iRiRiRiR .....21

RReq


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Current Electricity


In the process of calculation of resistance of a conductor if the integral elements arechosen is in such a way that same current passes through all the elements one after theother, then the resistance of elements are in series and the equivalent resistance of theconductor is given by

dRReq

In series combination, potential is to be divided. The potential difference across a resistorRisiRV

as iremains same in all the resistances connected in series, thus

RV therefore potential difference across R1 is

VR

RV

11

Parallel :When two terminals of an element are directly connected to the two terminals of the

other element, the two elements are called to be connected in parallel. The essentialcondition for the parallel connection is that the elements should bear the same potentialdifference.

in

i2

i1

I

E

R2

R1

Rn

niiiI .....21

neq R

E

R

E

R

E

R

E .....

21

RReq

11

In the process of calculation of resistance of a conductor if the integral elements are

chosen is in such a way that each element bears same potential difference and currentflows simultaneously in all the elements, then the resistance of elements are in paralleland the equivalent resistance of the conductor is given by

dRReq11

In parallel combination, current is to be divided. The current through a resistor is

RVi

as Vremains same across all the resistances connected in parallel, thus

Ri 1

therefore current through R1 is

i

R

Ri

1

11

1

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MARATHON-1016

Current ElectricityStar :When more than two elements branch out from the same node the network is said to be

star network. Here the node is a point in the circuit where more than two branches of

the circuit meet.

Delta :

When the three resistors form a triangular shape the network is said to be a deltanetwork. The delta network can be the part of a complete circuit.

C

RA RB

BDelta networkStar network

A

RCA RBC

RAB

Rc

BA

C

Conversion of delta network in to star network :Converting the delta network in to star network changes the complicated network in tosimple series and parallel connections which can easily be solved.If the two networks are equivalent then between any pair of terminals they must showthe same resistance. Thus

CABCAB

CABCABBA

RRR

RRRRR

..... (1)

CABCAB

ABCABCCB

RRR

RRRRR

..... (2)

CABCAB

BCABCA

AC RRR

RRR

RR

..... (3)

A

C

RA RB

B

RcRCA RBC

RABBA

C

Adding (1), (2) and (3)

CABCAB

ABCACABCBCABCBA

RRR

RRRRRRRRR 22

CABCAB

ABCACABCBCABCBA

RRR

RRRRRRRRR

.. (4)

Subtracting (1), (2) and (3) from (4) one by one we get

CABCAB

CAABA

RRR

RRR

,

CABCAB

BCABB

RRR

RRR

and


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Current Electricity


CABCAB

CABCC

RRR

RRR

Remembering these formulae is very easy. The delta you want to convert in to star, fix astar inside the delta. Now value of resistance of a branch of the star is product of the twoneighbor of this in delta divided by the sum of delta.

The reverse of the above is also possible and that is called star to delta conversion. Butat this stage we may not need to use them.

Wheatstone bridge :A network containing five elements connected as shown in the adjoining diagram iscalled wheatstone bridge. Bridge is said to be balanced if

SR

QP

b

QP

a

W

SR

In this condition the points a and b are at same potential. In a balanced bridge the

resistance Whas no role to play. The net resistance of the bridge remains independent of

Win a balanced bridge.


Calculate the equivalent resistance of the following network between A and B.

A B

P= 2

r

Q= 6

R= 5 S= 15

Solution :Converting the delta in to star in the given circuit

A B

q

p

BA r

15 5

62

rr

7

2

r7

10

r

r

7

5

q

p

6

15

r

r

7

2

r

r

7

5

r7

10

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MARATHON-1018

Current Electricity

Thus

157

56

7

2

157

56

7

2

7

10

r

r

r

r

r

r

r

r

rReq

rr

rr

rReq

714728

10520428

7

10

rr

Req

77

28040

7

40.

It is important to note that the equivalent resistance of this circuit is independent of the

resistance r. In fact this is a special configuration of the resistance network calledbalanced Wheat-Stone bridge. The Wheat-Stone bridge is formed by connecting fiveelements in the configuration shown in the problem and the bridge is said to be balancedif the ratio

S

R

Q

P .

In this condition the nodes p and q are at the same potential. Since the result isindependent of the resistance r directly connected between the equipotential points p

and qthus it can be noted that the equipotential points in a circuit can either be shortcircuited or open circuited, or they can be directly connected to any known or unknownelement, the net resistance of the network remains unaltered.

Q=6P=2

q

pBA

R=5 S=15

P= 2

r

Q= 6

R= 5S= 15

A B

q

p

P= 2 Q= 6

R= 5 S= 15

A B

q

p


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Cube solution :

Twelve resistance wires of equal resistance r are connected to form the structure of acube. The equivalent resistance is to be calculated between the pair of terminals for face

diagonal points (A, B), body diagonal points (A, C) and the points across one of the edgeof the cube (A, D).

B

C

A

D

1

2

3

4

Fig. 1

7

40eqR

7

40

28

160

208

208eqR

7

40

21

156

7

52eqR


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Current Electricity


The structure as shown in the diagram can be simply converted in two dimensionalfigure as shown bellow.

Terminal Ais a common terminal while irrespective of the other terminal (B, Cor D) we

can see the clear symmetry about the line AB. Therefore points 1 and 2 are

equipotential and this is also true for points 3 and 4 separately.

BCDA

BCDA

3

4

2

1

1, 2

3, 4

Fig. 2 Fig. 3

Now depending on the pair of terminals (AB, ACor AD) the circuit can be reduced toany one of the following

2r

23r

2r

CA

2r

23r

2r

2r

2r

BA

23r

23r

DA

2r

2r

r

r2r 2r

2r

Fig. 4(c)Fig. 4(b)Fig. 4(a)

AB:Solving the balanced wheat-Stone bridge

rReq 43 .

AC:Solving the unbalanced bridge using Star-Delta method we get the equivalent resistance

rReq 65 .AD:For this pair of terminals the circuit reduces to such a figure which can simply be

evaluated using series and parallel rules

rReq 127

Ladder networks :When a small resistive network is repeatedly added in length to form a long chain likenetwork, it is called a ladder network. The equivalent resistance of, infinite laddernetwork or the finite ladder network whose equivalent resistance is independent of the

number of units in the network can easily be calculated. For this except first link the

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MARATHON-1020

Current Electricitynetwork is replaced by the equivalent resistance which is to be calculated and theidentity is solved.


Figure shows an infinite circuit formed by the repetition of the small link, consisting of

resistance R1 and R2. Find the resistance of the circuit between the points A and B.

B

A

R2

R1

R2

R1R1

R2

R1

R2

R1

R2

Solution :The network can be reduced to the equivalent circuit as in the adjacent diagram.

B

A

R2

R1

ReqReq

2

2

1RR

RRRR

eq

eq

eq

221122

RRRRRRRRR eqeqeqeq

02112 RRRRR eqeq

2

411 1

1

2 R

R

RReq

Resistance of resistive networks with line of symmetry :

i) If a resistive network is symmetrical about the line joining the terminals then all thepoints in one half of the network are equipotential with corresponding points in the otherhalf of the network. Hence to reduce network one half can be folded on to the other halffor the equipotential points to get short-circuited.

ii) If a resistive network is symmetrical about a line perpendicular to the line joining theterminals then all the points lying on the line of symmetry are equipotential. They caneither be short-circuited or open-circuited depending on the requirement with outaffecting the equivalent resistance of the network.


Find the equivalent resistance of the resistive network of twelve identical resistances of

resistance r each, between terminals A and B.


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Current Electricity


A B

Solution :

The resistive network is symmetrical about the line joining terminals Aand Bthereforethe lower half can be folded over the upper half, hence

A B A Br r

2r

2r

2

r

2r

A Br r

2

r 2

r

4

r

2

r

4r

P

Q

Fig. 1(c)Fig. 1(b)Fig. 1(a)

The resistive network is still symmetric about the line PQ, the line perpendicular to the

line joining the terminals AB. Therefore point Pand Qare equipotential, hence joining P

and Qwe get

A Br r

2r

2r

4

r 4

r

P A Br r

2

r

2

r

4r

4r

2

r 2

r

or

Fig. 1(d) Fig. 1(e)

Solving both the figures 1(d) and 1(e) with the help of simple series and parallel rules weget

RReq 5

4

Or we can convert the middle delta in to star in figure 1(b) giving a balanced wheat stonebridge.

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A Br r

2r

2r

6

r

6r

Fig. 1(f)

r3

2

BA

r r

Fig. 1(g)

r3

2

6r


20/34

MARATHON-1022

Current ElectricitySolving the balanced wheat stone bridge we get

RReq 54

Resistance of a conductor having regular shapes :

Resistance of a solid conductor of regular shape can be calculated using integration. Ifthe resistance of integral elements come in series then

dRReqIf the resistance of integral elements come in parallel then

dRReq11


Find the resistance of a thick uniform pipe of length l, with inner radius R1, outer radius

R2 and resistivity of material .

Solution :

Let us consider an integral element in the form of an annular disc of thickness dxat a

distance xfrom one end of the pipe as shown

x dx

i

Resistance of the element for the shown current is

2122 RRdx

dR

Since all the elements are in series, hence

2122 RRdx

R

l

dxRR

R0

2

1

2

2

2122 RR

lR

Instead if we consider an integral element in the form of a thin co-axial cylinder of

thickness drof radius ras shown


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21/34

23MARATHON-10

Current Electricity


i

Resistance of the element for the shown current is

rdr

ldR

2

Since all the elements are in parallel, hence

lrdr

R

21

2

1

21R

R

rdrlR

2

212

1

2

2 RR

lR

2122 RRlR

Cell :Is a device which provides electrical energy to the system by pumping the charge fromnegative end to the positive end. It does not store any charge in itself.

deal cell Real cell

rE

Emf of the cell :Is equal to the potential difference between the terminals of the cell when no current isdrawn from the cell. In fact it is the energy provided by cell in circulating unit positivecharge throughout the circuit including cell itself.

Combination of identical cells :

Here we are considering the combination of identical cells, each of emf Eand internal

resistance r.

Series :

Ifnidentical cells are connected in series, the equivalent resistance of the circuitRnrR eq

and net emf in the circuit

,nEE eq

R

rE

rE

rE

n

R

nrnE

therefore current in external resistance

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22/34

MARATHON-1024

Current Electricity

nRr

E

Rnr

nE

R

Ei

q

q

e

eif R> r

R

Ei

which is independent of number of cells. Thus when external resistance is too large,parallel combination of cells cant increase current through it.

Mixed (battery) :

If total Nidentical cells arranged as mrows each containing ncells in series, are in turnconnected in parallel then the equivalent resistance of the circuit

Rrm

nReq

and net emf

,nEEeq

R

rE

rE

rE

rE rE rE

rE

rE

rE

n

m

R

nrnE

nrnE

nrnE

m

R

mnrnE

therefore current in the external resistance


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23/34

25MARATHON-10

Current Electricity


Rr

mn

nEi

mRnr

mnEi

mRnr

NE

i For the maximum current through the external resistor we should have net internalresistance of the battery equal to the external resistance. Thus for maximum current

through R

Rrm

n

mRnrthen the maximum current through external resistance

r

mEi

2max

Combination of non-identical cells : The emf of the equivalent cell is open circuit potential difference between Aand B. Let

loop current be ithen

A B

r1

E1

r2E2 i

A B

r1E1

r2E2

21

21

rr

EEi

22 irEVV BA

221

212 r

rr

EEEVV BA

21

1 2 2 1 1 2

1 21 2

1 1eq

EE

E r E r r r Er r

r r

and

21

21

rr

rrreq

If the cells are connected with opposite polarity then E2 is replaced by E2.

Let us generalize the above relation

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24/34

MARATHON-1026

Current Electricity

I

R

r1E1

r2

E2

rnEn

m

n

n

n

eq

rrr

rE

rE

rE

E1...11

...

21

2

2

1

1

and

r

rE

Eeq 1

SOLUTION OF CIRCUITS :There are three popular methods of solving a simple circuit as bellow.

Kirchoff's law - I (Loop rule) :is also known as Kirchoffs voltage law based on law of conversation of energy. The

algebraic sum of the changes in potential around any closed path is zero.potential drop

current i


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P.drop

+E E + qC

iR dt

diL

dt

diL

To apply this rule to solve a circuit, we will follow following steps.

Identify all elementary loops in the circuit.

Mark loop currents for all the loops clock wise (i1, i2, ... etc.).

Traverse the loop, one at a time, in the direction of loop current taking potentialdrop positive.

Solve simultaneous equations (thus formed in terms of unknown currents).

If potential difference between two points A and B in the circuit (VA VB) is to be

calculated then starting from A follow any path of circuit to reach B calculatingpotential drop positive.

Representative example 5(a) :

Calculate the current in 10resistance in the circuit shown in the adjoining diagram.

8 V

4 V

2 V

30

20

10

C+

EE Ri

Li +

current decreasing

L

ncreasing

+i


25/34

27MARATHON-10

Current Electricity


Solution :

As we see, we have two elementary loops. Let us assume the loop currents to be i1 and i2as shown in the diagram. Writing equations for the potential drop for loops (I) and (II) weget

4 V

i2i1

202 V

10

8 V

30

0210420 121 iii ..... (1)22030 21 ii

122 204308 iii ..... (2)125020 21 ii

Solving (1) and (2) for i1 and i2 we get,i1

7

55 amp.

Kirchoff's law-II (Junction rule) :

is also known as Kirchoffs currentlaw. It is based on law of conservation of charge. Ata junction (node) in a circuit, the incoming current equals the outgoing current. In other

words, the algebraic sum of the currents entering any junction point in a circuit is zero.In an electrical circuit, in a single branch the order of components (resistance,cell, capacitor, inductor etc.) does not affect the performance of the circuit, orcurrent in that branch. Therefore the same type of components can be collected atone place in a branch in an electrical circuit.


Identify all nodes (points where more than two branches meet) in the circuit. Collectcomponents branch wise.

Absolute potential of a point in a circuit can never be calculated, only the potential

difference between two points of circuit can be found. Mark one node as referencenode (consider its absolute potential to be zero) so that the potential of other nodescan be written with respect to this reference node. For this purpose generally thatnode is selected with which maximum number of branches are connected. Mark

other nodes at unknown potentials (V1, V2, ... etc.).

Take one node at a time (except reference node). Consider the currents coming tothat node in all the branches connected to that node.

Mark potentials at both the ends of the resistance of each branch connected to thatnode.

Calculate currents in the resistance of all the branches connected to that node goingtowards to the node and put there sum zero.

Solve the simultaneous equations (thus formed in terms of unknown potentials).

Representative example 5(b) :


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26/34

MARATHON-1028

Current Electricity8 V

4 V

2 V

30

20

10

Solution :As we see, we have two nodes. Let us assume the bottom node as the reference node.

Writing equations for the currents reaching at the upper node at potential V

2 V

10

V

8 V

30

20

4 V

(V+ 8)

(V2)

4 V

0 V0 V

0

30

80

20

4

10

20

VVV

11

8V volts

Thus current in 10 resistance is

10

211

80 i

55

7i amp.

Note that in this method we need only one equation to solve the circuit. If more numberof node potentials can directly be calculated due to absence of resistance in the branch,this method is simpler to use.

Superposition rule :Is very general rule. The statement of the law is sum of the effects is equal to the effectof the sum. In an electrical circuit the sum of the effects of individual cells is same asthe effect when all of them present simultaneously.


One cell is considered present at a time all the other cells are short-circuited. In thisprocess if sum internal resistance is there in the cells it will remain connected asit is.

The desired effect is calculated due presence of single cell in the circuit.

The processes is repeated for all the cells.

All the effects are added algebraically to get the result.

Representative example 5(c) :



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27/34

29MARATHON-10

Current Electricity


8 V

4 V

2 V

30

20

10

As we see, we have three cells in the circuit. Therefore we need three circuits. In each

circuit we will calculate the current in 10 resistance using simple series and parallelrules.

103020 10

4 V

3020

10

8 V

3020

2i 3i 1i

2 V

55

5

3020

302010

21i

55

8

3010

301020

42

I 55

12

2010

201030

83

I

55

6

3010

3022 Ii

55

8

2010

2033 Ii

Net current in 10 resistance

55

7

55

8

55

6

55

5i

Finding potential difference between two points on a circuit : Solve the entire circuit (find currents in all branches) using any of the three methods

discussed above (preferably Kirchoffs voltage law).

Move from point A to B following any path of your choice on the circuit taking

potential drop positive. The final result gives BA VV .

Branch of the circuit containing capacitor :

We talk about two kinds of behavior of a capacitor in a dc circuit.First, if the capacitor is connected in the circuit for sufficiently long time and thecurrents through various parts of the circuit (and voltages across various elements of thecircuit) are not changing with time. The capacitor is fully charged (refer to page 6 of

capacitance) and current through the branch containing capacitor is zero (that why sometimes it is also said that when a capacitor is fully charged it starts behaving as opencircuit element), then the circuit is said to be in stable state and the behavior of thecapacitor as steady state behavior of the capacitor.

If the capacitor is given in a circuit in steady state either potential difference across it orcharge on it or energy stored in it is to be calculated. To calculate any of these potentialdifference across the capacitor is to be found. In steady state current through capacitoris zero hence there is no current flowing through the branch of the circuit containingcapacitor. To find potential difference across a capacitor in steady state of the circuit

Ignore the entire branch in which capacitor is present (consider it removed from the

circuit).

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28/34

MARATHON-1030

Current Electricity Solve rest of the circuit (find currents in all branches) using any of the three methods

discussed above (preferably Kirchoffs voltage law).

Consider the branch containing capacitor again their in the circuit at its place.

Since no current flows through the branch containing capacitor hence if it containsany resistance there no potential drop across this resistance therefore this resistancehas no role to play in steady state of the circuit.

Move from one terminal of the capacitor (A) to the other terminal (B) following anypath of your choice on the circuit taking potential drop positive. The final result gives

, the potential difference across the capacitor.BA VV

Second, if the current through the branch containing capacitor is nonzero. In thissituation capacitor is either charging (acquiring charge and gaining potential differenceacross it) or discharging (loosing charge as well as potential difference across it). In thissituation currents through various elements of the circuit are changing with time andthe circuit is said to be in transient state and the behavior of the capacitor as transientbehavior of capacitor.

Leakage current through capacitor :R

+

There exists a small, constant current through a capacitor (here we will consider parallel

plate capacitor for example) in steady state due to conductivity of dielectric between the

plates. The equivalent of a charged capacitor filled by a dielectric of dielectric constant k

and resistivity is as shown in the adjoining diagram. The resistance offered by the dielectric between the plates of a parallel plate capacitor

with plate area Aand separation dis

A

dRC

hence the leakage current is

d

AV

R

Vi C

C

C

This is the constant current keeps flowing through parallel plate capacitor if thecapacitor is maintained in stable state.

Growth and decay of current in RCcircuit (transient behavior) :

Charging of capacitor :

A

E

K

RC

i

If at a certain time t, Cacquires a charge q, then from Kirchoff's voltage law

0 EiRCq differentiating


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29/34

31MARATHON-10

Current Electricity


001

dt

diR

dt

dq

Cas

idt

dq ,

RC

dt

i

di integrating

KRC

ti ln

At t= 0 the capacitor was uncharged, thus q= 0, therefore

R

Eii 0

0ln iKPutting this value ofKwe get

0lnln i

RC

ti

RCt

eii

0

RC

t

eR

E

i

Time constant ofRCcircuit :

Dimensions of quantity RC are the dimensions of time as the power of e should be a

dimension less quantity. RC = is called the time constant of circuit. The obvious

definition of time constant of circuit is the time constant of RCcircuit is that time in

which the current reduces by a factor ofe. It is also defined as the time in which currentbecomes zero, if it keeps decaying with the initial rate.

i0

i

t

RC

t

eR

Ei

RCt

eCR

E

dt

di

2

CR

E

dt

di

t

2

0

.

Equation of line passing through (0, i0) with slope

CR

E2

is

020

t CR

Eii , putting i= 0, we get

R

Et

CR

Ei

20

RCt

Potential difference across the capacitor :

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30/34

MARATHON-1032

Current Electricityqmax, E

q, Vc

t

C

qVc

iREVc

ReR

EEV RC

t

c

RC

t

c eEV 1

Charge acquired by the capacitor :

cCVq

RC

t

eCEq 1

RC

t

eqq 1max

Energy stored in the capacitor :2

21

cCVU

2

2 12

1

RCt

eCEU

The rate at which energy stored in the capacitor changes :

RC

t

RC

t

eRC

eCEdt

dUP

1012

RC

t

RC

t

eeR

EP 1

2

RC

t

RC

t

eeR

EP

22

Discharging of capacitor :

Similarly we can calculate these quantities at the time of discharging a charged capacitor

charged to an initial potential difference ofEvolts. Let the charge on capacitor at a time t

is q(at t= 0, q= CE) then from Kirchoffs loop law

0C

qiR (1)

01

dt

dq

Cdt

diR


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31/34

33MARATHON-10

Current Electricity


A

K

RC

+

i

as the capacitor is discharging rate of decrement of charge on the capacitor is same asthe current, hence

i0

t

q, Vc

qmax, E

dt

dqi therefore

0C

i

dt

diR

dtRCidi 1

KRC

ti ln (2)

initially

therefore from equation (1), at t= 0 the current isCEq0i

R

Ei 0

from equation (2)

Ki 0ln 0putting value of K in equation (2)

0lnln iRC

ti

RC

t

e

R

Ei

Similarly potential difference across the capacitor :

iRC

qVc

RCt

ceEV

charge on the capacitor :

cCVq

RCt

eCEq

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RCt

eqq

max


32/34

MARATHON-1034

Current ElectricityEnergy stored on the capacitor

2

21

CCVU

RCt

eCEU2

2

21

Galvanometer :Is basically a current meter for very small currents (of the order of mA or some times

even less). The current which gives full scale deflections in galvanometer is, say ig and

the resistance offered by the galvanometer is G. Since it gives full-scale deflection for very

small currents thus it can be used, as ammeter of desired range with small modificationsonly.

Conversion of Galvanometer in to ammeter and voltmeter :

Ammeter :It is an instrument used to measure currents. It is put in series in that branch of acircuit in which current is to be found out. An ideal Ammeter has zero resistance,

otherwise . To convert a galvanometer in to ammeter, a precalculated shunt

(parallel) resistance is connected with the galvanometer. If the desired range of the

ammeter is Ithen shunt current is

0AR

giI

GiSiI gg

g

g

iI

GiS

Voltmeter :It is an instrument to find the potential difference across two points in circuit. It is

essential that the resistance RV of a voltmeter be very large compared to the resistance ofany circuit element, which the voltmeter is connected. Otherwise, the meter itselfbecomes an important circuit element and alters the potential difference that is

measured. For a good voltmeter RV >> R or VR . For an ideal voltmeter RV = . If

a series resistance RV is needed to convert the galvanometer in to a voltmeter which gives

full scale deflection at potential Vthen

Vg RGiV

Gi

VR

g

V

1. Potentiometer : It is an arrangement used to measure open circuit potentialdifference between two points in a running circuit. In this arrangement a constantcurrent is established in a straight uniform wire of given length. The current generates a

uniform potential gradient between the two ends of the wire. When a potential differenceis to be measured first the potentiometer is calibrated.

E0

A


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33/34

35MARATHON-10

Current Electricity


To calibrate the potentiometer a cell of Known emf is balanced on the potentiometer

wire. For this positive terminals of the two cells are connected to common point and the

other terminal of is brought in contact through a galvanometer on the potentiometer

wire with the help of a sliding contact. The galvanometer has zero in the middle andgives deflection on both the sides depending on the direction of current flowing through

it. The length on wire from common point at which potentiometer gives no deflection iscalled calibration length . With this potential gradient in the wire is found in known

terms.

CE

CE

Cl

E0

A

rEC

G

L

lC

potential gradient in the wire = C

C

E

l

Finding emf of unknown cell : After calibrating the potentiometer the above process is

repeated with cell of unknown emf, the balancing length this time is l. The potentialgradient must be equated to know the emf of unknown cell

C

C

EEl l

CC

lE El

Finding emf of unknown cell : To compare the emf of two cells calibration of

potentiometer is not required. Two cells are balanced on the same wire turn by turn

following above process. If their balancing lengths are l1 and l2 respectively then

E0

A

r2E2

G

l1

r1E1

l2

1

2

11 C

C

lE E l and 22 C

C

lE E l hence

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34/34

MARATHON-1036

Current Electricity1 1

2 2

E lE l

Finding internal resistance of unknown cell : To find internal resistance of anunknown cell again calibration of potentiometer in not required. First the cell is balanced

following the above process against the wire length l1 (say). Then the cell is short

circuited with the help of a known resistance R which changes balancing length to asmaller value l2. Then

E0

A

rE

G

l2

l1

2

1

R

1C

C

lE E

l

and

2C

C

lV E

l

1

2

lEV l

Now potential difference Vacross the cell when it is short circuited with the help of aresistance Ris

V E ir iR

V

V E rR

1r

V ER

1E

r RV

1

2 1

l

r R l

Documents

01 35 Current Electricity Study