Assignment 1(Special 2006)

(Solution)

CIRCUIT THEORY (PHY301)

MARKS: 40

Due Date: 23/09/2006

Q.1. Find equivalent resistance R for the following circuit. Draw and label the circuit

diagram of each step otherwise you will lose your marks because diagram of each step

contains marks.

SOL:

6and 12 are in parallel so their equivalent will be

12x6/12+6=72/18=4

4 and 1 are in series their sum will be

4+1=5

Now 5 and 6 are in parallel

6x5/11=30/11=2.72

2.72 and 8 are in parallel its equivalent will be

2.72x8/10.72=21.98/10.72=2.035

2.035 and 10 are in series so sum will be

10+2.035=12.035that is equivalent resistance

Q.2. Determine the unknown voltage and current of each of following circuit.

SOL: In fig. (a) The current IT will take the path of least resistance and since the short

circuit condition at the end of network is the least resistance path, all the current will pass

through short circuit. The voltage across the network will be same as that across the short

circuit and will be zero volts, as shown in fig.

For the network in fig( b) The circuit is open circuit , therefore no current will pass

through the network. The voltage across the resistors must therefore be zero volts, as

determined by ohm’s law (V=IR=OV) with the resistor simply acting as a connection

from supply to the open circuit. The result is that the open circuit voltage will be E=20v

as shown in fig. below.

Q.3. In the network given below the power absorbed by the 7

resistance is 252W. Find VS also find out the value of IS , IB and IA.

SOL:

P=252W= IB

2 x 7 So, IB

2 =252/7 IB

2 = 36 IB= 6A ----- (A)

Where by ohm’s Law VX = 6 x 7 = 42 V IA = 42 /42 IA = 1A ------ (B)

From (A) and (B) we have IS = IA + IB = 7A VS = 5 x 7 + VX = 35 + 42

VS = 77 V

Q.4. Briefly answer the following questions.

a) What is difference between inductance and capacitance?

Ans: Resistance offered by an inductor in an circuit is called inductance whereas

capacitance is the resistance offered by a capacitor in a circuit.

b) Give three differences between voltage and current.

The volt is the measure of the work needed to move an electric charge.

When potential difference between two charges forces a third charge to move the charge

in motion is called current.

Charge in motion is called current or we can say current is constant flow of electrons.

Potential difference is V for voltage Volt is unit of potential difference.

Ampere is the unit of current the charge in motion.

Voltage is measured by voltmeter while current is measured by ammeter.

c) Give one advantage and one disadvantage of parallel connection.

Advantage: Voltage remains same in parallel connections. With parallel circuits we

have more control over the circuit if we want to switch current at one side

Disadvantage: In parallel connection current divides .If you have multiple power

sources, the power stays at the same voltage as that of the single power source. But in

parallel circuits increasing the number of output devices does not increase the resistance

the way it does in series circuits

d) Give two safety factors to consider for preventing an electric shock.

Ans: Use lather glows while working with electric circuits.

Use non conductor shoes(Plastic or rubber) while working with electric circuits and don’t

touch wires with bare hands, also u can use tester for checking electricity in circuit.

e) Briefly define the following, giving its unit and symbol potential difference,

voltage divider, & conductance.

Ans: Potential difference

Potential refers to the possibility of doing work. Any charge has the potential to do the

work of moving another charge, by either attraction or repulsion. When we consider two

unlike charges, they have a difference of potential. The unit of potential difference is volt

and its symbol is V.

Voltage divider

Any series circuit is voltage divider. The IR drops are proportional part of the voltage.

Special formulae can be used for voltage division as short cuts in calculations. The

voltage formula gives the series voltage even when the current is unknown.

Formula is stated below

V = VT * R / RT

Conductance

It is reciprocal of resistance. The lower the resistance the higher the conductance.

Its unit is siemens and symbol is G

Assignment 2(Special 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 30

Due Date: 02/10/2006

Q.1. What is the difference between Loop, Mesh and Super Mesh? What is the difference between Node and Reference node? Elaborate your point of view with the help of diagrams. (Note: You must Draw complete circuit diagram.) Sol.

Mesh: A Mesh is a special kind of loop that does not contain any other loops within it. For example , the network given below contains two meshes defined by the path a-b-e-f-a, b-c-d-e-b

Loop: The path a-b-c-d-e-f-a is a loop not a mesh.

If two Mesh currents are passing through a current source, then form a Super Mesh or Loop around that source. Current source will put a constraint on the values of the two Mesh currents passing through it, giving an extra equation.

NODE: is the junction of two or more than two elements or we can say that it is the point of connection between circuit. Reference node: The reference node is commonly called the ground since it is assumed to have zero potential. The choice of a reference node is completely arbitrary, but the node with the largest number of components or voltage source connected to it is usually most convenient. Super node: A node which emerges as a result of combination of two ordinary nodes around a voltage source is called a super node. Constraint or coupling equation describes a super node mathematically, instead of writing equations individual ordinary nodes of the super node. The difference in potential between the two nodes is equal to voltage source between two nodes. Now to write constraint equation first write the node which is towards +ve terminal of voltage source then subtract the node which is towards -ve side of voltage source then equal this difference to value of voltage source.

Q.2.

In the circuit given below just label the direction of conventional current and electronic current in each mesh also find out the polarity across each resistance with respect to each current.

(Note: You will only label the above diagram.) Sol.

The motion of positive charge, in the opposite direction from electron flow, is considered as conventional current. Electricity was known of long before Benjamin Franklin. It was not understood very well, but it was known of. Scientists knew there were two kinds of electric charge. They knew there was electric current. Scientists believed that the opposite charges moved similarly in opposite directions. They defined one as positive and one as negative. They defined current to be in the direction of the positive charges. Later, they learned of their mistake. Only the negative charges move freely in conductors. Electrons had been defined as the negative charges. Current had been defined "backwards". It was too late to redefine all of electrical physics, so the inconvenience holds to this day. The direction that the electrons move are opposite the direction that current points. Because of how electricity works, it isn't much of a problem. Negative charge moving to the left through a wire has the same effect as positive charge moving to the right. So long as the total charge in the wire (protons and electrons) remains balanced, no trouble occurs.

Conventional Current assumes that current flows out of the positive terminal, through the circuit and into the negative terminal of the source. This was the convention chosen during the discovery of electricity. They were wrong!

Electronic Flow is what actually happens and electrons flow out of the negative terminal, through the circuit and into the positive terminal of the source.

Both Conventional Current and Electron Flow are used by industry. Many textbooks are available in both Electron Flow and Conventional Current formats.

In fact, it makes no difference which way current is flowing as long as it is used consistently. The direction of current flow does not affect what the current does.

Q.3. If IO = 4mA in the network given below. Find IS.Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol.

VCD = -12 + (I0)(4 kΩ) VCD = -12 + (4 mA)(4 kΩ) VCD = -12 + 16 VCD = 4 Volts

VCD

I3 = 2 kΩ

4 V I3 = 2 kΩ I3 = 2 mA By KCL

I2 = I0 + I3

I2 = 4 mA + 2 mA I2 = 6 mA VAC = I2(1 kΩ) VAC = (6 mA)(1 kΩ) VAC = 6 × 1 VAC = 6 Volts VAB = VAC + VCD

VAB = 6 V + 4 V VAB = 10 Volts

VAB

I1 = 10 kΩ

10 V I1 = 10 kΩ I1 = 1 mA By KCL IS = I1 + I2

IS = 1 mA + 6 mA IS = 7 mA

------ Good Luck -----

Assignment 3(Special 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 08/10/2006

Q.1. Find VO in the circuit given below. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value. (Note: You must Draw complete circuit diagrams.) Sol.

VAB = (12 mA)(2 kΩ)

VAB = 24 Volts

Current in branch CD

VAB

IO = 4 kΩ

24 V

IO = 4 kΩ

IO = 6 mA VCE = (Io)(2 kΩ) VCE = (6 mA)(2 kΩ)

VCE = 12 Volts

VCE

I0 = 3 kΩ

12 V

I0 = 3 kΩ

I0 = 4 mA

Q.2. Use nodal analysis to find VO in the network given below. Identify and label each node otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. Sum of all the currents leaving the junction = sum of all the currents entering the junction V0 – 12 V0 – (-9) V0 – (-6) = + 12 k 18 k 12 k + 4 k V0 – 12 V0 + 9 V0 + 6 = + 12 k 18 k 16 k

V – 12 V + 9 V + 6 - - = 0 12 k 18 k 16 k 12[V0 - 12] – 8[V0 + 9] – 9[V0 + 6] = 0

144 k 12[V0 - 12] – 8[V0 + 9] – 9[V0 + 6] = 0 12V0 - 144 – 8V0 - 72 – 9V0 - 54 = 0 -5V0 = 270 V0 = -54 V

Q.3. Use nodal analysis to find IO in the network given below. Identify and label each node otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value. Sol. Constraint Equation V1 – V2 = 4000Ix … (A) V1 V1 V2

+ + = 4 mA 2k 2 k 4 k 2V1 + 2V1 + V2

= 4 mA 4 k

2V1 + 2V1 + V2 = (4 k)(4 mA)

4V1 + V2 = 16 … (B) Here V1

Ix = 2 k Substituting the value of Ix in (A) V1

V1 – V2 = 4000 2000 V1 – V2 = 2V1

V1 – V2 - 2V1 = 0 V1 = -V2 … (C)

Substituting the value of V1 in (B) 4[-V2] + V2 = 16 -4V2 + V2 = 16 -3V2 = 16 V2 = -5.334 Volts

V2 I0 =

4 k

-5.334 V I0 =

4 k I0 = -1.333 mA Substituting the value of V2 in (C) V1 = -[-5.334 V] V1 = 5.334 Volts

Q.4. Use Mesh analysis to find IO in the given network. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol.

Sol.

Mesh: I1 = 2 mA I3 = 4 mA Loop I2: According to KVL 2 [I2 – I1] + 4I2 + 6000[I2 – I3] = 12 2I2 – 2I1 + 4I2 + 6I2 – 6I3 = 12 12I2 – 2I1 – 6I3 = 12 Substitutig the values of I1 & I3

12I2 – 2[2 mA] – 6[4 mA] = 12 12I2 - 4 – 24 = 12 12I2 = 40 I2 = 3.333 mA = I0

------ Good Luck -----

Assignment 4(Special 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 06/11/2006

Q.1. Use Superposition to find IO in the network given below. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value. Sol. Only Voltage sources are acting

Series combinations = 2 kΩ + 2 kΩ = 4 kΩ = 2 kΩ + 2 kΩ = 4 kΩ

According to KVL Sum of voltage rise = sum of voltage drop 4000I + 6 = 12 4000I = 6 I = 1.5 mA I = I0' ------------(A) Only Current sources is acting

Parallel combination

2 kΩ × 2 kΩ

= 2 kΩ + 2 kΩ

4 k × k

= 4 k

= 1 kΩ

According to current divider rule

2 k I2k = × 2 mA 2 k + 2 k I2k = 1 mA I2k = I0" I0 = I0' + I0" From (A)

I0 = 1.5 mA + 1 mA I0 = 2.5 mA

Q.2. Use Source transformation to find VX in the network given below. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol.

Voltage division rule;

VX = (-160x12)/12+8+20 = -48V

Q.3. Use Source transformation to find VO in the network given below. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol. 6 V I = 3 k I = 2 mA 5 V I = 2 k I = 2.5 mA

R1 = 3 kΩ R2 = 2 kΩ R3 = 2 kΩ

R1R2R3

= R1R2 + R2R3 + R3R1

(3 k)(2 k)(2 k) =

(3 k)(2 k) + (2 k)(2 k) + (2 k)(3 k)

12 k × k × k =

6 k × k + 4 k × k + 6 k × k

12 k × k × k = 16 k × k = 0.75 kΩ

2 mA 2 mA 2.5 mA

= 2 mA + 2 mA – 2.5 mA = 4 mA – 2.5 mA = 1.5 mA

1.5 mA

1 kΩ

1.5 mA 0.75 kΩ 1 kΩ

According to current divider rule

0.75 k

I 2k = × 1.5 mA

2 k + 0.75 k I2k = 0.409 mA According to ohm’s Law V0 = (I2k)(1 k) V0 = (0.409 mA)(1 k) V0 = 0.409 V

------ Good Luck -----

Assignment 5(Special 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 19/11/2006

Q.1. Use Thévenin’s theorem to find IO in the network given below. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol. Only Voltage sources are acting

VTH =? Using Mesh analysis: From Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 6[I1 – I2] + 2[I1 – I2] + 1I1 = 12 6I1 – 6I2 + 2I1 – 2I2 + 1I1 = 12 9I1 – 8I2 = 12 From Mesh I2 :

I2 = 2 mA 9I1 – 8[2] = 12 9I1 – 16 = 12 9I1 = 28 I1 = 3.111 mA

According to ohm’s Law V6k = (I1 – I2)(6 k) V6k = (3.111 mA – 2 mA)(6 k) V6k = (1.111 × 10-3)(6 × 10+3)

V6k = 6.666 V VTH = 12 V – 6.666 V VTH = 5.334 V

THEVENIN’S EQUIVALENT:

5.334 I0 = 2 k + 2 k I0 = 1.333 mA

Q.2. Use Norton’s theorem to find VO in the network given below. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol.

Using Mesh analysis: From Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2000I1 = 12 I1 = 6 mA ISC = I1 – I2

Here I2 = -2 mA

ISC = 6 mA – [-2 mA] ISC = 8 mA

RTH = 2 kΩ

NORTON’S EQUIVALENT: According to current divider rule: 2 k I2k = × 8 mA 2 k + 2 k I2k = 4 mA = I0

According to ohm’s Law: V0 = I0(2 k) V0 = (4 mA)(2 k) V0 = 8 V

Q.3. For the circuits shown in the Figure below using ideal diodes, find the values of the voltages and currents indicated, mention the units of each derived value. (Comment & Hint: Easy to see if diode is conducting or not by inspection) Sol.

(A) Diode On by inspection,

Diode ≡ short circuit ⇒ V=-3V I = 3-(-3)/10 =0.6mA

(B) Diode Off by inspection,

Diode ≡ Open circuit ⇒ V=+3V I = 3-(3)/10 =0

(C) Diode On by inspection,

Diode ≡ short circuit ⇒ V= 3V I = 3-(-3)/10 =0.6mA

(D) Diode Off by inspection,

Diode ≡ Open circuit ⇒ V=-3V I = 3-(3)/10 =0

Q.4. A diode, for which the forward voltage drop is 0.7V at 1.0mA and for which n=1 is operated at 0.5V what is the value of the current? Use this problem to derive equation V2 – V1 = n VT ln(I2 / I1) where Assume VT= 25mV [Hint: Diode equation is i = Is (eV/nV

T -1) (Ignore "-1" in diode eqn.)] Sol. i = I

s(e

v/nVT – 1) ≅ i = I

s(e

v/nVT)

Find I when n =1 and V=0.5V

ln i = ln Is(e

v/nVT )

ln i = ln Is + ln(e

v/nVT )

put n= 1 ln i = ln I

s + V/VT

Consider (i1, v1), (i2, v2) ln i1 = ln I

s + V1/VT ---------------(A)

ln i2 = ln Is + V2/VT ---------------(B)

Subtracting (A) and (B) we have ln i1 - ln i2 = lV1/VT - V2/VT

ln (i1/i2) = 1/VT (V1- V2) VTln (i1/i2) = V1 - V2

Now (i1, v1) = (1mA , 0.7V) ln (1mA/i2) = (0.7- 0.5) /25mV therefore, i2 = 335mA

------ Good Luck -----

Assignment 6(Special 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 29/11/2006

Q.1. A circuit shown in the figure below has values of N1= 12, N2=1, V1=150Vac, and RL= 8.2 kΩ. What is the peak load current for the circuit? Sol. The input voltage is given an rms value. This value is converted to a peak value as follows: V

1(pk) = V

1(rms)/0.707

= 150Vac

/0.707

V1(pk)

= 212.2 Vpk

Now, the load voltage and current are found, after fining peak voltage, as V

2(pk) = N2/N1 V1(pk)

= (1/12)(212.2 Vpk

)

V2(pk)

= 17.7Vpk

Finally, the load voltage and current and current values are found as: V

L(pk) = V

2(pk) – V

F

= 17.7 – 0.7 V

L(pk)= 17V

pk

and the current will be I

L(pk) = V

L(PK)/R

L

= 17 Vpk

/8.2k

IL(pk)

= 2.1mApk

Q.2. A bridge rectifier is fed by a 22 Vac transformer. Determine the dc load voltage and current for the circuit when it has a 1.9kΩ load. Sol.

With the 22Vac

rated transformer, the peak secondary voltage is found as

V2(pk)

= 22/0.707

=31.12Vpk

The peak load voltage is now found as

VL(pk)

= V2 – 1.4

VL(pk)

=29.72Vpk

The dc load voltage is found as V

ave = 2V

L(pk)/Π

=59.44/Π V

ave =18.93 V

dc

Finally the dc load current is found as I

ave = V

ave/R

L

= 18.93/1.9k I

ave = 996.31µA

Q.3.

In a particular BJT, the base current is 7.5µA, and the collector current is 400µA. Find β and α for this device. Sol.

β = 400µA/7.5µA = 53.33

And α=IC/IE = 400µA/407.5µA = 0.982

Q.4. Find the values of β that correspond to α values of 0.99, 0.995, and 0.999. Sol.

α = 0.99 ⇒ β = α/(1−α) = 99

= 0.995 ⇒ β = α/(1−α) = 199 ∴∆ β >> ∆ α

= 0.999 ⇒ β = α/(1−α) = 999

Q.5. A PNP transistor operates with an emitter-to-collector voltage of 5V, an emitter current of 10A, and

VEB =0.85V. For β =15, what base current is required? What is IS for this transistor. Sol. IC = α IE ----------- (A)

Where α = β/(β+1) = 15/16 so put in (A) IC = 15 x 10 /16 9.375A

∴ IB = IC /β = 625mA IE = IS eV

EB /VT

⇒ = 9.375/e0.85/25mV = 1.607 x 10-14

------ Good Luck -----