Circuit Theory - Solved Assignments - Spring 2006

Assignment 1(Spring 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 05/05/2006

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• Labeled and draw each circuit diagram, other wise you will lose your marks. Write each step of the calculation to get maximum marks.

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Q.1. For the circuit shown in the figure below, all the resistors are given in Ohms; Find the total resistance RT in the following circuits. 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.

Q.2. You are given the circuit Use Current Division Rule directly to find I30.

Sol. From Current division rule we have

I30 = (10 x 20) / (20 + 30) = 200/50 V I30= 4A

Q.3. In the network given below find the voltage VX .

Sol. By Ohm’s Law

I1 = 30/15 = 2A Vab = 2x 25 = 50V I2 = 50 / 50 = 1A

From KVL (entering current equal to leaving current ) I3 = I1 + I2 = 3A VX = 15I3 + Vab

= 45 + 50 VX = 95 V

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



Due Date: 16/05/2006

Q.1. Many years ago a string of Christmas tree lights was manufactured in the form shown in Fig. A. Today the lights are manufactured as shown in Fig. B. Is there a good reason for this change? Circuit diagram: Figure A Figure B Sol. In figure (A) if any of the light is not working, the other lights will not lighten up on turning the switch on. While in figure (B) every light is directly connected to the source of current. So, if any of the light is not working it will have no effect on the other lights.

Q.2. Find Ix, Iy, and Iz in the network given below.

Sol. Fist of all we will label the diagram, For node 1: According to KCL: Iz + 4 mA = 2 mA Iz = -2 mA For node 2: According to KCL: Iz + 12 mA + Iy = 0 Substituting the value of Iz

-2 mA + 12 mA + Iy = 0 Iy = - 10 mA For node 3: According to KCL: 12 mA = 3 mA + Ix

Ix = 9 mA

Q.3. Use nodal analysis to find both VO and V1 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. Fist of all we will label the diagram,

KCL equation at Node’V1’ will be, Sum of all the currents leaving the junction = sum of all the currents entering the junction V1 V1 – V2

+ - 2 mA + 4 mA =0 3 k 6 k V1 V1 – V2

+ = -4 mA + 2 mA 3 k 6 k V1 V1 – V2

+ = -2 mA 3 k 6 k 2V1 + V1 – V2

= -2 mA 6 k 2V1 + V1 – V2 = (6 k)(2 mA) 3V1 – V2 = (6 × 10+3)(2 × 10-3) 3V1 – V2 = 12 × 10+3-3

3V1 – V2 = 12 × 100

3V1 – V2 = 12 × 1 3V1 – V2 = -12 …………..… (A) KCL equation at Node’V2’ will be, V2 V2 V2 – V1

+ + + 2 mA =0 12 k 4 k 6 k V2 V2 V2 – V1

+ + = -2 mA 12 k 4 k 6 k V2 + 3V2 + 2[V2 – V1] = -2 mA 12 k V2 + 3V2 + 2[V2 – V1] = (12 k)(-2 mA) V2 + 3V2 + 2V2 - 2V1 = (12 × 10+3)(-2 × 10-3) 6V2 - 2V1 = -24 × 10+3-3

V1 - 3V2 = 12 × 100

V1 - 3V2 = 12 × 1 -V1 + 3V2 = -12 ………………..… (B) -V1 = -12 - 3V2

Substituting the value of V1 in (A) 3[-(-12 - 3V2)] – V2 = -12 36 + 9V2 – V2 = -12 8V2 = -12 – 36 8V2 = -48 V2 = -6 Volts Substituting the value of V2 in (B) V1 -3(-6) = 12 V1 = 12 – 18 V1 =-6 Volts According to Voltage Division Rule:

2 k V0 = × V2

2 k + 2 k 2 k

V0 = × -6 V 4 k

V0 = -3 Volts

Q.4. Use nodal analysis to find both 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. Fist of all we will label the diagram,

Applying KCL at Super node V1 V1 – V3 V2

+ + - 2 mA = 0 4 k 2 k 2 k V1 V1 – V3 V2

+ + = 2mA 4 k 2 k 2 k V1 + 2[V1 – V3] + 2V2 = 2mA

4 k V1 + 2[V1 – V3] + 2V2 = (4 k)(2 mA) V1 + 2V1 – 2V3 + 2V2 = (4 × 10+3)(2 × 10-3) 3V1 - 2V3 + 2V2 = 8 × 10+3-3

3V1 – 2V3 + 2V2 = 8 × 100

3V1 – 2V3 + 2V2 = 8 × 1 3V1 + 2V2 – 2V3 = 8 ………………… (A) Constraint equation V2 – V1 = 12 …………………(B) V2 = 12 + V1

Substituting the value of V2 in eq. (A) 3V1 + 2[12 + V1] – 2V3 = 8 3V1 + 24 + 2V1 – 2V3 = 8 5V1 – 2V3 = 8 - 24 5V1 – 2V3 = -16 ………………………..(C) Applying KCL at node ‘V3’ V3 – V1 V3

+ 2 mA + =0 2 k 2 k

V3 – V1 V3

+ = -2 mA 2 k 2 k

V3 – V1 + V3

= -2 mA 2 k 2V3 – V1 = (2 k)(-2 mA) 2V3–V1 = (2 × 10+3)(-2 × 10-3) 2V3–V1 = -4 × 10+3-3

2V3–V1 = -4 × 100

2V3–V1 = -4 × 1 2V3 - V1 = -4 Volts ……………. (D) Solving (C) and (D) simultaneously we have V1 = -5V

Substituting the value of V1 in eq. (B) V2 + 5 = 12 V2 = 7 Volts V0 = V2 = 7 Volts

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



Due Date: 25/05/2006

Q.1. Find VO in the network given below. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. According to Voltage Division Rule

4 k V0 = × 12 V 4 k + 8 k

4 k V0 = × 12 V 12 k V0 = 4 Volts

Q.2. Use Mesh analysis to find Voltage VO 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.

Here I1 = 2 mA …………….… (A) Loop I2: According to KVL Sum of all the voltage drop = sum of all the voltage rise 1000[I2 – I1] + 2000[I2 – I3] + 1000I2 = 12 1000I2 – 1000I1 + 2000I2 – 2000I3 + 1000I2 = 12 -1000I1 + 4000I2 – 2000I3 = 12 -[1000I1 - 4000I2 + 2000I3] = 12 1000I1 - 4000I2 + 2000I3 = -12 Substituting the value of I1 from equation (A)

1000[2 mA] - 4000I2 + 2000I3 = -12 2 - 4000I2 + 2000I3 = -12 - 4000I2 + 2000I3 = -14 -[4000I2 - 2000I3] = -14

4000I2 - 2000I3 = 14 ……….… (B) Loop I3: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2000[I3 – I2] + 1000[I3 – I1] + 1000I3 = 6 2000I3 – 2000I2 + 1000I3 – 1000I1 + 1000I3 = 6 -1000I1 – 2000I2 + 4000I3 = 6 Substituting the value of I1 from equation (A)

-1000[2 mA] – 2000I2 + 4000I3 = 6 -2 – 2000I2 + 4000I3 = 6 –2000I2 + 4000I3 = 8 -[2000I2 - 4000I3] = 8

2000I2 - 4000I3 = -8 ……………..(C) Multiplying both sides of Equation (C) by 2 4000I2 - 8000I3 = -16 …………….… (D) Subtracting equation (D) from (B) 4000I2 - 2000I3 = 14 4000I2 - 8000I3 = -16 - + + 6000I3 = 30 I3 = 5 mA Substituting the value of I3 in equation (B) 4000I2 – 2000[5 mA] = 14 4000I2 – 10 = 14 4000I2 = 24 I2 = 6 mA According to ohm’s Law V0 = (I2)(1 k)

V0 = (6 mA)(1 k) V0 = (6 × 10-3)(1 × 10+3) V0 = 6 × 10-3+3

V0 = 6 × 100

V0 = 6 × 1 V0 = 6 V

Q.3. Use Mesh analysis to find Voltage VO 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. Technique # 1

From fig. (A) Constraint Equation I2 – I3 = 12 mA ……….… (I) Loop I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 12000I1 + 6000[I1 – I3] + 4000[I1 – I2] = 0 12000I1 + 6000I1 – 6000I3 + 4000I1 – 4000I2 = 0

22000I1 – 4000I2 – 6000I3 = 0 From equation (I) I3 = I2 – 12 mA 22000I1 – 4000I2 – 6000[I2 – 12 mA] = 0 22000I1 – 4000I2 – 6000I2 + 72 = 0 22000I1 – 10000I2 = -72 …………….… (II) Writing equation for Super Mesh According to KVL Sum of all the voltage drop = sum of all the voltage rise 6000[I3 – I1] + 4000[I2 – I1] + 8000I2 = 10Vx

6000I3 – 6000I1 + 4000I2 – 4000I1 + 8000I2 = 10Vx

-10000I1 + 12000I2 + 6000I3 = 10Vx

From equation (I) I3 = I2 – 12 mA -10000I1 + 12000I2 + 6000[I2 – 12 mA] = 10Vx

-10000I1 + 12000I2 + 6000I2 – 72 = 10Vx

-10000I1 + 18000I2 – 72 = 10Vx

We know from the above circuit diagram Vx = 4000[I1 – I2] -10000I1 + 18000I2 – 72 = 10[4000(I1 – I2)] -10000I1 + 18000I2 – 72 = 40000(I1 – I2) -10000I1 + 18000I2 – 72 = 40000I1 – 40000I2

-10000I1 + 18000I2 – 72 - 40000I1 + 40000I2 = 0 -50000I1 + 58000I2 = 72 -[50000I1 - 58000I2] = 72 50000I1 - 58000I2 = -72 ………………… (III)

Multiplying both the sides of equation (II) by –50 -1100000I1 + 500000I2 = 3600 ………………… (IV) Multiplying both the sides of equation (iii) by 22 1100000I1 - 1276000I2 = -1584 ………………… (V) Adding (IV) & (V) - 1100000I1 + 500000I2 = 3600 +1100000I1 - 1276000I2 = -1584

-776000I2 = 2016 I2 = -2.60 mA According to ohm’s Law V0 = (I2)(8 k) V0 = (-2.60 mA)(8 k) V0 = (-2.60 × 10-3)(8 × 10+3)

V0 = -20.8 × 10-3+3

V0 = -20.8 × 100

V0 = -20.8 × 1 V0 = -20.8 V

Technique # 2

I1 I2 I3

= + −

= + −

=

+ + + + − + − =

=

+ − + + + + − + − =

+ − = − − − − −

3 1 2

3 1 2

3

1 3 3 1 2 1 2

3

3 1 2 1 3 3 1 1 2

3 1 2

( )4

4 4 4

12

110 8 ( ) 4 ( ) 6 ( ) 0

12 ( )

10(4 4 4 ) 8 ( ) 4 ( ) 6 ( ) 0

52 58 50 0

FromthediagramwehaveVx I I I k

Vx kI kI kI

I mA

For MeshVx k I I k I I I k I I

Put I mA in A wehave

kI kI kI k I I k I I I k I I

I I I

=

+ − =

− = − − − − − − − − − − − −

− + − − + = − − − − −

=

− + − − + =

− + − − + =

3

1 2

1 2

2 1 2 1 3 2

3

2 1 2 1 2

2 1 2 1 2

( )

12 ( )

52(12) 58 50 0

29 25 312 (1)

26 ( ) 4 ( ) 12 0 ( )

12 ( )

6 ( ) 4 ( 12 ) 12 0

6 6 4 4 48 12 0

1

A

Put I mA in A wehave

I I

I I

For Meshk I I k I I I kI B

Put I mA in B wehave

k I I k I I mA kI

kI kI kI kI kI

− = − − − − − − − −1 20 22 48 (2)kI kI

( )= − = −

Ω= + Ω

= −= −

1 2

3 1

(1) 2

14.599 4.454

8 tan( )8

(12 14.599)820.88

Solving and simultenouslywehave

I mAand I mA

To find out thevoltage across K reis ceVo I I k

VoVo volts

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



Due Date: 30/06/2006

Q.1. Use Mesh analysis to find Current through each mesh in the given network also find out Voltage Vo. Identify and label each mesh and also show each step of calculation otherwise you will lose your marks. Draw the complete circuit diagram and also mention the units of each derived value.

Sol.

0 1 2

01

01

We can see from the figure that V = 2k(I + I ) -----------(A)

From Mesh 1:2V

I = 10002V

I = 1k

[ ]

[ ]

1 21

1 21

1 21

1 21

From (A) we have

2 2k(I + I ) I =

1k2 2kI + 2kI

I = 1k

4kI + 4kI I =

1k4k(I + I )

I = 1k

1 1 2

1 1 2

1 1 2

1 2

I = 4(I + I )

I = 4I + 4I

I - 4I = 4I

-3I = 4I

1 2

4 I = I --------(1)

3

2

2 1 2 1 2

2 1 2 1 2

1

Loop Equation for Current I

4kI + 4k(I + I ) + 2k(I + I ) = 6

4kI + 4kI + 4kI + 2kI + 2kI = 6

6kI + 10kI

:

2

1 2

1 2

1 2

= 6

2 (3kI + 5kI ) = 6

3kI + 5kI = 3

4Put I = - I in the above Eq. we have

3

2 2

2 2

2

4 3k - I + 5kI = 3

3 -4kI + 5kI = 3

1kI = 3

⎛ ⎞⎜ ⎟⎝ ⎠

2

2

I 3mA --------(B)

By putting the value of I in eq. (1) we have

4

=

Q.2. Apply Superposition to the network given below to find out VO. 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 Source is acting First we will replace current source by an open circuit, so we will have the circuit as shown below

From the above figure 6kΩ is in parallel with 6kΩ, so 6kΩ || 6kΩ = 3kΩ Thus the circuit will become as

Now from the circuit we see that 6kΩ is in series 3kΩ, so 9 6k+3k k= Ω The circuit will be as,

In the above diagram 6kΩ is in parallel with 9kΩ, so 6kΩ || 9kΩ = 3.6kΩ According to KVL Sum of voltage drop = sum of voltage rise 6000I + 3600I = 12 9600I = 12 I = 1.25 mA According to ohm’s Law

V0′ = (I)(6 k) V0′ = (1.25 mA)(6 k)

V0′ = 7.5 V Only Current source is acting Now we will replace voltage source by short circuit, so our circuit will be as

Re-arranging the above diagram as

According to current divider rule 6 k

I = × 2 mA 6 k + 6 k I = 1 mA According to ohm’s Law V3k = (I)(3 k) V3k = (1 mA)(3 k) V3k = 3 V V3k = VAB

V3k = V0″ V0 = V0′ + V0″ V0 = 7.5 V + 3 V V0 = 10.5 V

Q.3. Find IO in the network given below by using Source Transformation . 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.

For Mesh I: According to KVL Sum of all the voltage drop = sum of all the voltage rise 4I1 + 12[I1 – I2] + 12 = 0 4I1 + 12I1 – 12I2 = -12 16I1 – 12I2 = -12 … (A) For Mesh II: According to KVL Sum of all the voltage drop = sum of all the voltage rise 3I2 + 12[I2 – I1]+3I2 = 6 6I2 + 12I2 – 12I1 = 6 –12I1 + 18I2 = 6 … (B) Multiplying both sides of equation (A) by 3 48I1 – 36I2 = -36 … (I) Multiplying both sides of equation (B) by 2 –24I1 + 36I2 = 12 … (II) Adding equations (I) & (II) 24I1 = -24 I1 = -1 mA According to ohm’s Law V4k = (I1)(4 k) V4k = (-1 mA)(4 k) V4k = (-1 × 10-3)(4 × 10+3) V4k = -4 V V4k = VAB

I0 =VAB/6K = -4V/6K I0 = -0.667 mA

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



Due Date: 12/07/2006

Q.1. Use Source Transformation 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.

Technique I

Technique II

Parallel combination

12 kΩ × 6 kΩ

= 12 kΩ + 6 kΩ

72 k × k

= = 4 kΩ 18 k Source transformation: V = (2 mA)(3 k) V = 6 V Using mesh analysis: For Mesh I: According to KVL 6I1 + 4[I1 – I2] + 12 = 0 6I1 + 4I1 – 4I2 + 12 = 0

10I1 – 4I2 = -12 ----------(A) For Mesh II: According to KVL 12I2 + 4[I2 – I1] + 6 = 0 16I2 – 4I1 = -6 Rearranging -4I1 + 16I2 = -6 ------------(B)) Multiplying both sides of equation (A) by (4) 40I1 – 16I2 = -48 ----------(C) Adding equations (B) & (C) 36I1 = -54 I1 = -1.5 mA Substituting the value of I1 in equation (A) 10[-1.5] – 4I2 = -12 -15 – 4I2 = -12 -4I2 = 3 I2 = -0.75 mA I4k = I1 – I2

I4k = -1.5 mA – (-0.75 mA) I4k = -1.5 mA + 0.75 mA I4k = -0.75 mA According to ohm’s Law V4k = (I4k)(4 k) V4k = (-0.75 mA)(4 k) V4k = -3 V V4k = VAB

VAB

I0 = 6 k -3 V I0 = 6 k I0 = -0.5 mA

Q.2. Use a combination of Thevenin’s theorem and Superposition 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. (Note: You must solve this problem only using the techniques mentioned in the problem’s statement.)

Sol.

VTH =? Consider 12 V source:

Series combination: = 2 kΩ + 8 kΩ + 4 kΩ = 14 kΩ

Parallel combination:

6 kΩ × 14 kΩ

= 6 kΩ + 14 kΩ

84 k × k

= 20 k = 4.2 kΩ

According to Voltage divider rule: 4.2 k V4.2k = × 12 V 4.2 k + 3 k V4.2k = 7 V V4.2k = VAB

From fig. (c) According to Voltage divider rule: 10 k V10k = × VAB

10 k + 4 k 10 k V10k = × 7 V 14 k V10k = 5 V V10k = VTH’

Now consider 4 mA source:


3 kΩ × 6 kΩ =

3 kΩ + 6 kΩ

18 k × k = 9 k = 2 kΩ

Loop I2: According to KVL Sum of all the voltage drop = sum of all the voltage rise 12000[I1 + I2] + 4000I2 = 0 12000I1 + 12000I2 + 4000I2 = 0 12000I1 + 16000I2 = 0 Here I1 = -4 mA 12000[-4 mA] + 16000I2 = 0 -48 + 16000I2 = 0 16000I2 = 48 I2 = 3 mA According to ohm’s Law V8k = (I1 + I2)(8 k) V8k = (-1 mA)(8 k) V8k = -8 × 1

V8k = -8 V According to ohm’s Law V2k = I2(2 k) V2k = (3 mA)(2 k)

V2k = 6 V From fig. (g) VTH” = V8k + V2k

VTH” = -8 V + 6 V VTH” = -2 V VTH = VTH' + VTH" VTH = 5 V + (-2 V) VTH = 3 V RTH =? Parallel combination:

3 kΩ × 6 kΩ =

3 kΩ + 6 kΩ

18 k × k = 9 k = 2 kΩ Series combinations: = 2 kΩ + 4 kΩ = 6 kΩ = 2 kΩ + 8 kΩ = 10 kΩ Parallel combination:

10 kΩ × 6 kΩ =

10 kΩ + 6 kΩ

60 k × k = 16 k = 3.75 kΩ

THEVENIN’S EQUIVALENT:

According to Voltage divider rule: 4 k V0 = × 3 V 3.75 k + 4 k V0 = 1.548 V

Q.3. Find VO in the network given below using Thevenin’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. VTH =?

Using Mesh method: Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 6000I1 + 2000[I1 – I2] + 3 = 0 Here I2 = -1 mA 6000I1 + 2000[I1 – (-1 mA)] + 3 = 0 6000I1 + 2000I1 + 2 + 3 = 0 8000I1 + 5 = 0

I1 = -0.625 mA Ix = I1 – I2

Ix = -0.625 mA – [-1 mA] Ix = 0.375 mA According to ohm’s Law: V2k = Ix(2 k) V2k = (0.375 mA)(2 k) V2k = 0.75 V VTH = V2k + 1000Ix

Here Ix = 0.375 mA VTH = 0.75 V + 1000[0.375 mA] VTH = 0.75 V + 0.375 V VTH = 1.125 V ISC =?

Equation for supernode: V1 + 3 V1 V2

+ + = 1 mA 6 k 2 k 1 k V1 + 3 + 3V1 + 6V2

= 1 mA 6 k V1 + 3 + 3V1 + 6V2 = (1 mA)(6 k) 4V1 + 3 + 6V2 = 6 × 1 4V1 + 6V2 = 3 2V1 + 3V2 = 1.5 … (A) Constraint Equation: V2 – V1 = 1000Ix

V1

Ix = 2 k V1

V2 – V1 = 1000 2 k V2 – V1 = 0.5V1

V2 – V1 - 0.5V1 = 0 V2 – 1.5V1 = 0 V2 = 1.5V1

V1 = 0.667V2

Substituting the value of V1 in equation (A) 2[0.667V2] + 3V2 = 1.5 1.334V2 + 3V2 = 1.5 4.334V2 = 1.5 V2 = 0.346 V V2

ISC =

2 k 0.346 V ISC = 1 k ISC = 0.346 mA VTH

RTH = ISC

1.125 V RTH = 0.346 mA RTH = 3.251 kΩ

THEVENIN’S EQUIVALENT: According to Voltage divider rule: 2 k V0 = × 1.125 V 2 k + 3.251 k V0 = 0.428 V

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



Due Date: 24/07/2006

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. VTH =? 2000Ix - + + 6 kΩ 4 kΩ

- +

2 kΩ 3 mA VTH

6 V Ix

- +

-

2000Ix +

6 kΩ 2 kΩ 3 mA VTH

- +

6 V I1 I2 Ix

-

- +

Using Mesh method: Loop I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 6000I1 + 2000[I1 – I2] + 6 = 0 Here I2 = -3 mA 6000I1 + 2000[I1 – (-3 mA)] + 6 = 0 6000I1 + 2000I1 + 6 + 6 = 0 8000I1 + 12 = 0 I1 = -1.5 mA Ix = I1 – I2

Ix = -1.5 mA – [-3 mA] Ix = 1.5 mA According to ohm’s Law: V2k = Ix(2 k) V2k = (1.5 mA)(2 k) V2k = 3 V VTH = V2k + 2000Ix

Here Ix = 1.5 mA VTH = 3 V + 2000[1.5 mA] VTH = 3 V + 3 V VTH = 6 V

ISC =? 2000Ix V1 V2 6 kΩ 4 kΩ 2 kΩ 3 mA ISC

6 V Ix

- +

Equation for supernode: V1 + 6 V1 V2

+ + = 3 mA 6 k 2 k 4 k 2V1 + 12 + 6V1 + 3V2

= 3 mA 12 k 2V1 + 12 + 6V1 + 3V2 = (3 mA)(12 k) 8V1 + 12 + 3V2 = (3 × 10-3)(12 × 10+3)

8V1 + 3V2 = 24 ………… (A) Constraint Equation: V2 – V1 = 2000Ix

V1

Ix = 2 k V1

V2 – V1 = 2000 2 k V2 – V1 = V1

V2 – V1 - V1 = 0 V2 – 2V1 = 0 V2 = 2V1

V1 = 0.5V2

Substituting the value of V1 in equation (A) 8[0.5V2] + 3V2 = 24 4V2 + 3V2 = 24 7V2 = 24 V2 = 3.428 V V2

ISC = 4 k 3.428 V ISC = 4 k ISC = 0.857 mA VTH

RTH = ISC

6 V RTH = 0.857 mA RTH = 7 kΩ NORTON’S EQUIVALENT:

ISC = 0.857 mA 7 kΩ 4 kΩ

According to current divider rule: 7 k I4k = × 0.857 mA

4 k + 7 k I4k = 0.545 mA According to ohm’s Law: V4k = I4k(4 k) V4k = (0.545 mA)(4 k) V4k = (0.545 × 10-3)(4 × 10+3) V4k = 2.18 V = V0

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. 4 kΩ

+ 3 kΩ 4 mA 8 kΩ

12 V 6 kΩ 2 kΩ 4 kΩ V0

-

+ -

ISC =?

4 kΩ I4

3 kΩ 4 mA 8 kΩ

I3

12 V 6 kΩ I2 2 kΩ I1

ISC

+ -

Using Loop analysis: Loop I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 3000I1 + 6000[I1 – I2 – I4] = 12 3000I1 + 6000I1 – 6000I2 – 6000I4 = 12 9000I1 – 6000I2 – 6000I4 = 12 Here I2 = 4 mA 9000I1 – 6000[4 mA] – 6000I4 = 12 9000I1 – 24 – 6000I4 = 12 9000I1 – 6000I4 = 36 …………… (A) Loop I4: According to KVL Sum of all the voltage drop = sum of all the voltage rise 4000I4 + 6000[I4 + I2 – I1] = 0 4000I4 + 6000I4 + 6000I2 – 6000I1 = 0 10000I4 + 6000I2 – 6000I1 = 0 Here I2 = 4 mA 10000I4 + 6000[4 mA] – 6000I1 = 0 10000I4 + 24 – 6000I1 = 0 -6000I1 + 10000I4 = -24 ……………… (B) Multiplying both the sides of eq. (A) by 2 18000I1 – 12000I4 = 72 ………………… (C) Multiplying both the sides of eq. (B) by 3 -18000I1 + 30000I4 = -72 ……………………..… (D) Adding equations (C) & (D) 18000I4 = -12 I4 = 0 Loop I3: According to KVL Sum of all the voltage drop = sum of all the voltage rise 8000I3 + 2000[I3 – I2] = 0 8000I3 + 2000I3 – 2000I2 = 0 10000I3 – 2000I2 = 0 Here I2 = 4 mA 10000I3 – 2000[4 mA] = 0 10000I3 – 8 = 0 10000I3 = 8 I3 = 0.8 mA ISC = I4 + I3

ISC = 0 mA + 0.8 mA ISC = 0.8 mA

RTH =? 4 kΩ o.c 3 kΩ 8 kΩ

s.c 6 kΩ 2 kΩ RTH


3 kΩ × 6 kΩ =

3 kΩ + 6 kΩ

18 k × k = 9 k = 2 kΩ

4 kΩ 8 kΩ

2 kΩ 2 kΩ RTH

Fig. (I) Series combinations: = 2 kΩ + 4 kΩ = 6 kΩ = 2 kΩ + 8 kΩ = 10 kΩ

6 kΩ 10 kΩ RTH

Fig. (II) Parallel combination:

10 kΩ × 6 kΩ =

10 kΩ + 6 kΩ

60 k × k = 16 k

RTH = 3.75 kΩ

3.75 kΩ RTH

NORTON’S EQUIVALENT:

ISC = 0.8 mA 3.75 kΩ 4 kΩ

According to current divider rule: 3.75 k I4k = × 0.8 mA 3.75 k + 4 k I4k = 0.38 mA = I0

According to ohm’s Law: V0 = I0(4 k) V0 = (0.38 mA)(4 k) V0 = 1.54 V

Q.3. Find IO in the network given below. Using Linearity and the assumption that IO=1mA 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 Linearity


6 kΩ × 2 kΩ =

6 kΩ + 2 kΩ

12 k × k =

8 k = 1.5 kΩ

A 2 kΩ 3 kΩ

+ -

64 V 6 kΩ 1.5 kΩ

B Fig (A) Series combination = 1.5 kΩ + 3 kΩ = 4.5 kΩ

2 kΩ 64 V 6 kΩ 4.5 kΩ


6 kΩ × 4.5 kΩ =

6 kΩ + 4.5 kΩ

27 k × k

= 10.5 k

= 2.571 kΩ

2 kΩ 64 V 2.571 kΩ

+ -

+ -

According to voltage divider rule

2.571 k V2.571k = × 64 V 2.571 k + 2 k V2.571k = 35.997 V V2.571k = VAB From fig. (A) According to voltage divider rule

1.5 k V1.5k = × VAB

1.5 k + 3 k Here VAB = 35.997 V

1.5 k V1.5k = × 35.997 V 1.5 k + 3 k V1.5k = 11.999 V V1.5k = V6k = V2k

V6k

I0 = 6 k 11.999 V I0 = 6 k I0 = 2 mA Using assumption

I4 I2

2 kΩ I3 3 kΩ 64 V 6 kΩ 2 kΩ 6 kΩ I0 I1 V6k = (I0)(6k)

+ -

V6k = (1 mA)(6k) V6k = 6 V From fig. (a) V6k = (I0)(6k) V6k = (2 mA)(6k) V6k = 12 V Hence

I0 = 2 mA

Q.4. Determine the currents I1, I2, and ID2 for 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.

The applied voltage is such as to run both diodes on, as noted by the resulting current direction in the network given below. Note the use of the abbreviated notation for “on” diodes and that the solution is obtained through an application of technique applied to dc series-parallel networks.

I1=VT/R1 = 0.7V/3.3k

=0.212mA Applying Kirchhoff’s voltage law around the indicated loop in the clock wise direction yields, -V2+E-VT1-VT2=0 V2 = E-VT1 -VT2

= 20V-0.7V-0.7V= 18.6V I2 = V2/ R2 = 18.6V/5.6k = 3.32mA

At the bottom node A ID2= I2-I1=3.32ma-0.212mA=3.108mA I

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



Due Date: 02/08/2006

Q.1. A “1-mA diode” (i.e. one that has VD =0.7V at iD = 1mA) is connected in series with a 200-Ω resistor to a 1.0-V supply.

(a) Provide a rough estimate of the diode current you would expect. (b) If the diode is characterized by n=2, estimate the diode current more closely using

iterative analysis. Sol.

(a) iD= (1-0.7)/0.2 = 1.5mA

(b) Iterative analysis gives vD=0.7V At iD=1mA # 1 v=0.7V iD= (1-0.7)/0.2 = 1.5mA

#2 ∴ i= Is eV/n VT n=2 and VT = 0.025 i2 /i1 = e (v2-V1)/0.05

thus v2= v1+0.05 ln i2/i1

∴ for i=1.5mA V=0.7 + 0.05 ln (1.5/1) = 0.720 V & iD= (1-0.720)/0.2 = 1.4mA #3 V=0.720+ 0.05 ln (1.4/1.5) = 0.716 V & iD= (1-0.716)/0.2 = 1.42 mA #4 V=0.716+ 0.05 ln (1.42/1.4) = 0.716 V & iD= (1-0.716)/0.2 = 1.42 mA

Q.2. A bridge rectifier is fed by an 18Vac transformer. Determine the dc load voltage and current for the circuit when it has a 1.2kΩ load. Sol.

With the 18V

ac rated transformer, the peak secondary voltage is found as

V2(pk)

= 18/0.707

=25.46Vpk

The peak load voltage is now found as

VL(pk)

= V2 – 1.4

VL(pk)

=24.06Vpk

The dc load voltage is found as V

ave = 2V

L(pk)/Π

=48.12/Π V

ave =15.32 V

dc

Finally the dc load current is found as I

ave = V

ave/R

L

= 15.32/1.2k I

ave = 12.76 µA

Q.3. For each of the circuits shown in the figure below, find the emitter, base, and collector voltages and currents. Use β = 30, but assume |VBE| = 0.7V independent of current level.

Sol. From fig (A) we have VB = 0V VE = VB+0.7 =0.7V IE= (3-VE) / 1 = (3-0.7)/1 = 2.3mA

IC = α IE = 30/31 x 2.3mA = 2.23mA VC = -3 + 1xIC = -3 + 2.23 = -0.77V IB = IC/ β = 2.23 / 30 = 0.0743mA From fig (B) we have VB = 3V VE = VB+0.7 =3.7V IE= (9-VE) / 1.1 = (9-3.7)/1.1 = 4.82mA

IC = α IE = 30/31 x 4.82mA = 4.66mA VC = IC x 0.56 = 2.62V IB = IC/ β = 4.66/ 30 = 0.155mA

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

Documents

Circuit Theory - Solved Assignments - Spring 2006