(Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 02/08/2007

Q.1. Find I1, I2, I3.

Draw the circuit diagram of each step otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol.

Identifying Node 1,2 and 3.

For Node 1 2A – I1+ 12A = 0 I1=14A

Assignment 1

For Node 2 -12A + I2 + 14A = 0 I2 + 2A = 0 I2 = -2A For Node 3 -14A + I3 + 4A = 0 I3- 10A = 0 I3 = 10A

Q.2.

Determine I1 to I5 . Draw the circuit diagram of each step otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. For finding I solve the Circuit.

1Ω || 2 Ω

= 1*2/1+2 = 2/3 = 0.67 Ω

4Ω || 0.67 Ω

= 4*0.67 / 4+0.67 = 2.68 / 4.67 = 0.57 Ω

0.57 Ω is in series with 3 Ω = 0.57 + 3 = 3.57 Ω Now find total current through the circuit Itotal= V / R = 40 / 3.57 Itotal = 11.20A From the given circuit we have

Where Itotal = I1= 11.20A To find I2 and I3 we will use current divider rule and their source current will be I1

I2= (R0.67Ω / R4Ω + R0.67Ω ) x I1 I2= (0.67 / 4.67) x 11.20 I2= 1.61A I3= (R4Ω / R4Ω + R0.67Ω ) x I1 I3= (4 / 4.67) x 11.20 I3= 9.60A To find I4 and I5 , once again we will use current divider rule and their source current will be I3

I4= (R2Ω / R2Ω + R1Ω) x I3 = (2 / 3) x 9.60 I4= 6.4A I5= (R1Ω / R2Ω + R1Ω) x I3 = (1 / 3) x 9.60 I5= 3.2A

Q.3. Find Req. Draw the circuit diagram of each step otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. The two 12kΩ are in parallel As 12kΩ || 12kΩ gives = 144 / 24 = 6kΩ At this particular point most of the students commit the following mistake and redraw the circuit as This technique is wrong because in the above circuit we displace the out put terminal which is connected at Node A form its original position and connect it at C. The correct circuit will be redrawn as The 6kΩ is in series with the top 6kΩ gives = 6 + 6 = 12kΩ This 12kΩ is parallel with 4kΩ = 48 / 16 Req= 3kΩ

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

Assignment 2(Summer session 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 08/08/2007

Q.1. Find V10. Draw the circuit diagrams if necessary otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. 20Ω and 5Ω are in series. So the combined effect will be equal to 20Ω+5Ω= 25Ω

1MΩ and 1MΩ are in parallel so same voltage 35v drops across each resistance. Now 35v divides to 10 and 25(sum of 20 and 5) which can be find by voltage division rule.

We know the voltage division formula

110

1 2

RV V

R R

⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠

10

10

10

1035

10 25

1035

35

10

V

V

V Volts

⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

=

Q.2.

Find IA Draw the circuit diagrams if necessary to elaborate your point of view otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. We can redraw the given circuit as

R3 =10Ω

R4 =4Ω

R2=30Ω

IA

AR1=40Ω

IB

IS

Is =10A

Since 10Ω are in series with 4 Ω so 10 + 4= 14 Ω and the circuit can be redrawn as follow

R3 =14Ω

R2=30Ω

IA

AR1=40Ω

IBIs =10A

IS

As the current Is passing through the R1 resistor is 30 A. At the node A current Is divides into IA and IB, so by current division rule,

2

3 2

2 3

( )

10 30 14

by putting these values in (A) we have,30

1014 3012044

6.81A

A s

s

A

RI I A

R R

we know I A R and R

I x

= − − − − −+

= = Ω =

=+

=

=

Ω

Q.3. First Identify and label each node in the network. Use nodal analysis to find voltage VO in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. Circuit can be redrawn as,

At node V0:

0 0 0 2

0 0 0 2

2 0

300

1 2 44 120 2 0

7 120 0 (

V V V V

V V V V

V V A

− −+ + =

− + + − =

− + = − − − − − − − )

At Super Node As the V2 and V3 made the super node so its equation will be ,

( )

( )

3 2 0

3 2 0

3 2 0

3 2 0

3 016 4

48 4 0

48 4 4 0

4 4 48 0

V V V

V V V

V V V

V V V B

−− + =

− + − =

− + − =

+ − − = − − − − − − − − − −

Now coupling equation of Super node will be as

( )2 3 0

3 2 0

2

2

V V V C

and V V V

− = − − − − −

= −

Put this value of V3 in (B), we have

2 0 2 0

2 0

02

2 4 4 48 0

5 6 48 0

6 48( )

5

V V V V

V V

VV I

− + − − =

− − =

+= − − − − −

Put this (I) in (A) we have,

00

0 0

0

0

0

6 487 120 0

5

6 48 35 600 0

29 648 0

29 648

64822.34

29

VV

V V

V

V

V V

+⎛ ⎞ − + =⎜ ⎟⎝ ⎠

+ − + =

− + =

− = −

= =

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

Assignment 3 (Summer session 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 23/08/2007

Q.1. Calculate V0 & IO. Draw the circuit diagrams if necessary otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol.

From output loop we have

Vo= 50Io x 20 x103 = 106Io ----------- (A)

From Input loop we have

3x103+ 4x103Io-Vo/100 =0 -------------- (B)

Solving (A) and (B) we have

IO = 0.5µA

And VO= 0.5V

Q.2.

Sol: Nodal equation at Vo will be

0 0 0

0 0 0

0 0 0

0

0

0

0

12 10 04 6 2

3( 12) 2 6( 10) 012

3 36

8

2 6 60 011 96 011 96

.7

961

2721

v

v v v

v v v

v v vvv

v

v

− −+ +

=

=

− + + −=

− + + − =− =

=

=

Alternative method

Labeling circuit for super node

Constraint equation becomes,

V0-V1=10 -----------------(A)

OR

V1 =V0 - 10

KCL equation for the super node is ,

0 0 1

0 0 1

0 1

12 04 6 2

3 36 2 6 012

5 6 36 0 (

V V V

V V V

V V B

−+ + =

− + +=

+ − = − − − − − − − − − )

Putting the value of the V1 in the (B)we have

( )0 0

0 0

0

0

5 6 V 10 36 05 60 6V 36 011 96 0

96 8.7211

VVV

V Volts Answer

+ − − =

− + − =− =

= =

Q.3.

First Identify and label each mesh in the network. Use mesh analysis to find voltage V1, V2 in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify mesh and labeled them,

From Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2(I1 –I2)+ 2(I1 – I3) = 12 2I1 - I2 – I3 = 6 ----------------- (A) From Mesh I2: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2(I2 –I1) + 2I2 +2(I2 – I3) = 0 -I1 +3I2 – I3 = 0 ----------------- (B) From Mesh I3: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2(I3 –I1) + 2(I2 – I3) +2I3= 0 -I1 -I2 +3I3 = 0 ----------------- (C) Solving (A) (B) and (C) simultaneously I1 = 6A, I2 = 3A and I3 = 3A And V1=2I2 = 6V V2=2I3 = 6V

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

Sol. First we will identify mesh and labeled them, From Mesh I1: I1 = -2A ---------(i) From Mesh I2: According to KVL Sum of all the voltage drop = sum of all the voltage rise 1(I2 –I1)+ 2(I2 – I4) + 9 + 4I2 = 0 I1 +7I2 – 2I4 = -9 Put I= -2A from (i) we have 7I2 – 2I4 = -11 ----------------- (ii) From the circuit we observe that a current source of 4A is sharing by Mesh I3 and Mesh I4 so Super Mesh will develop here

Now redraw the circuit for Super Mesh equation Super Mesh Eq: -24 + 4I3 + 3I4 +1I4 +2(I4 –I2)+ 2(I3 – I1) = 0 -2I2 +6I3 + 6I4 = 20 ---------------- (iii) Constraint equation: I4-I3=4 ----------- (iv) Solving equation (i) ,(ii), (iii) and (iv) simultaneously we have, I2 = -11/20 = -0.55 A I3 = -17/40 = -0.425 A I4 = 143/40 = 3.575 A From the circuit we know that Io= I1-I2 =-2-(0.55) = -1.45A

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

Assignment 4(summer session 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 30/08/2007

Q.1. Find the terminal voltage Vab by

(a) Superposition (b) Nodal Analysis

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.

(a) Super position Only Current source is acting -10x2 +3Vab1+Vab1=0 4 Vab1=20 Vab1=5V

Only Voltage source is acting -4V + +3Vab2+Vab2 = 0 4Vab2 = 4 Vab2 = 1 By Superposition Vab = Vab1 + Vab2 = 5+1 = 6V (b) By Nodal Analysis

At Node Vab : (Vab + 3Vab -4)/10 -2 = 0 Vab + 3Vab -4 - 20 = 0 4Vab = 24 Vab = 6V

Q.2. Apply Source Transformation on the circuit given below to find I. 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. Working circuit using source transformation is a little like a chess or checker game, different people will make different moves to get to the same end. What I see is making a current to voltage to current transformation to the left of a-b then makes a voltage to current transformation to the right of c-d. Then continue current source to one net current source, combine resistors to one resistor then use current division to find I. First Step to the left of a-b

Second Step to the right of c-d

Apply Current division rule I = 1x 5/(5+4) = 5/9 A =0.555A

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

Assignment 5 (summer session 2007) (SOLUTION)

CIRCUIT THEORY (PHY301) MARKS: 35

Due Date: 11/09/2007

Q.1. Find VO in the network given below using Norton’s theorem 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: First step: Replacing RL with a short circuit to find IN. Here RL is 6k resistor.

Second step: Finding IN using nodal analysis

Isc =I1 + I2

For node 1 V1/3k + (V1 – 6)/6k + V1/8k +4m=0

8V1 +4V1 – 24 +3V1 +96 =0

15V1+72=0

or V1= -4.8V

For node 2 V2/2k + V2/4k -4 =0

2V2 +V2 – 16 =0

3kV2 = 16

V2 =16/3

V2= 5.33V

Now I1 = V2/4k

= 5.33 /4k I1 = 1.33 mA

I2=V1/8 I2 = -4.8/8k

=-0.6mA So, IN = I1+ I2

IN = (1.33 -0.6)

IN = 0.73mA

Third step: Calculating RN To calculate RN we will short circuit all voltage sources and open the current sources

. For RN

3k|| 6k = 2k 2k is in series with 8k = 2k + 8k = 10k 4k is in series with 2k = 4k + 2k =6k 10k ||6k = 10 x6/16 RN =60/16

RN = 3.75k

Fourth Step:

After calculating IN and RN, re-inserting the load resistance RL in the circuit in parallel

RN and considering the IN current source parallel with these two resistances.

To calculate V0

I0 = 0.73m x 3.75k x 1/(6k+3.75k)

= 0.280mA V0 = 6k x 0.280m

V0 = 1.68 Volts

Q.2. Find the Thevenin equivalent circuit for the network in the shaded area between A, B. Draw and label the circuit diagram of each step and also mention the units of each derived value.

First step: removing load resistance between point A,B

Second step: Finding Vth Vth is the voltage drop across point A,B of removed resistance. Since terminal A,B and resistance 6Ω are in parallel so the circuit can be redrawn as

Voltage drop across 6Ω will be the Vth which can be find out by using voltage divider formula for series resistance of 6Ω and 4Ω Vth=8x6/6+4=48/10=4.8v Third step: Finding Rth To find the Rth we short circuit the voltage source Here 2Ω is short circuited its effect will be ignored in the circuit Now 6Ω and 4Ω are in parallel as terminal A,B exists between them so Rth=6x4/6+4=2.4Ω Therefore Thevenin equivalent circuit will be Q.3 what will be the source voltage Vs if 6.7v drops across resistor having germanium diode in series in the given fig. Sol: Source voltage (Vs) can be find using the formula VS=VR+Vf Because diode is germanium so 0.3v is dropped across it for its forward biasing s VS=6.7+0.3 VS=7v

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

Assignment 6(summer session 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 16/09/2007

Q.1. 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?

Sol. For

I = Is (e v/nVT

-1) ≅ Is e v/nVT

Given I=1mA at V=700mV and n=1 find I when V=500mV

V2-V1=nVT ln (I2/I1)

I2/I1 = e

v2 –v1/nVT

I2 =I1 e v2 –v1/nVT

I2 =1mA e 500 –700/1x25

=0.335µA =335nA

Q.2. For the circuit shown in the figure below, using the constant voltage-drop (VD = 0.7V) diode model, find the voltage and currents indicated.

Sol.

(A) Diode FB by inspection,

I = 3-(-3 +0.7)/10 = 530 µA =0.53mA

Diode FB , so replace by 0.7V source ⇒ V=3 – (0.53mA) (10k)= -2.3V

(B) Diode RB by inspection,

Diode ≡ Open circuit ⇒ V=+3V I = 0 (same as ideal diode)

(C) Diode On by inspection, 0.7V drop

I = 3- 0.7) –(-3)/10 = 530 µA =0.53mA

⇒ V= -3 + (10k)(0.53mA)= 2.3V

(D) Diode Off by inspection,

Diode ≡ Open circuit ⇒ V=-3V I = 0 (same as ideal diode)

Q.3.

Measurements on the circuits given below produces labeled voltages as indicated. Find the value of β for each transistor.

Sol. (A) ic/ib= β = 2/0.215 =9.3

(B)

ic / ib = β = 9.9/0.1 =99

(C)

β + 1= ie / ib = 3x 100/3.3 β = (300/ 3.3) - 1 = 89.9

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