ETN2B_Chapt3_Homework1

Circuit Theorems

Tutorial

Prepared by

Lecturer: Dr. Khmaies Ouahada

Course: ETN2B, 2010

B2 Lab 113Email: [email protected]

University of Johannesburg, South AfricaDepartment of Electrical and Electronic Engineering Science

P.O. Box 524, Auckland Park, 2006

1 Circuit Theorems

1.1 Question no 1

Given V1 = 10V , V2 = 20V , R1 = 6Ω, R2 = 1Ω, R3 = 2Ω, R4 = 3Ω, R5 = 10Ω.

a b

c d

R1 R3

R2 R4 R5

V1

V2

I+

−

− +

1. Draw the circuits corresponding to each voltage sources alone based on the super-

position theorem.

2. Calculate the current I by applying the superposition theorem.

1.2 Question no 2

Given V1 = 10V , V2 = 20V , R1 = 6Ω, R2 = 4Ω, R3 = 5Ω, R4 = 3Ω, R5 = 2Ω.


position theorem.

1-1

1. Circuit Theorems

R1 R3

R2

R5

V1

V2

I

+

−

−+

R4


1.3 Question no 3

Given V1 = 12V , V2 = 6V , R1 = 4Ω, R2 = 8Ω, I = 6A.

R2R1I

+ − − +

V1 V2

IR


position theorem.

2. Calculate the current IR by applying the superposition theorem.

1.4 Question no 4

Given V = 20V , I1 = 10A, I2 = 5A, R1 = 2Ω, R2 = 1Ω, R3 = 3Ω and R4 = 4Ω.

1-2

1. Circuit Theorems

R3R2

I2 R4

IR4

R1

I1

V

+ −


position theorem.

2. Calculate the current IR4 by applying the superposition theorem.

1.5 Question no 5

Given V1 = 70V , V2 = 50V , R1 = 4Ω, R2 = 20Ω, R3 = 2Ω and R4 = 10Ω.


position theorem.


1-3

1. Circuit Theorems

R2

I

R1

V1

+ −

V2R3

R4

2I1− +

+

−

I1

1.6 Question no 6

Given V1 = 6V , V2 = 10V and R1 = R2 = R3 = 10Ω. Find the voltage V by applying

the superposition theorem.

R1

R2

V1

+

−

R3

V2

+

−

V

1.7 Question no 7

Given V1 = 3V , V2 = 2V , V3 = 1V , R1 = 2Ω, and R3 = 1Ω.

1-4

1. Circuit Theorems

R1 R3

R2V1

V2

I2

+

−

−+

V3

−

+

I3I1 a

b

1. Find the current I1 using the preposition theorem if R2 = 0.

2. Find the current I2 using the preposition theorem if R2 = 0.

3. Find the current I1 using the preposition theorem if R2 = 1Ω.

1.8 Question no 8

Given V = 90V , R1 = 8Ω, R2 = 6Ω, R3 = 8Ω and R4 = 8Ω, R5 = 4Ω and R6 = 5Ω.

Calculate the Thevenin voltage and resistance at the terminals ab of the circuit in the

figure below.

R2

R4 R6V1

−

R1

R5

+

a

b

R3

1-5

1. Circuit Theorems

1.9 Question no 9

Given V1 = 100V , V2 = V3 = 50V , R1 = 10Ω, R2 = 2Ω, R3 = 5Ω and R4 = 2Ω and

R5 = 3Ω. Calculate the current I2 in the resistor R3 of the circuit shown in the figure

below by using the Thevenin’s theorem.

R2

R3 R4−

R1

R5

+

V3

− +

V2

− +

V1

I1

I2

I3

1.10 Question no 10

Given V1 = 20V , V2 = 9V , V3 = 50V , V4 = 10V , R1 = 10Ω, R2 = 1Ω, R3 = 20Ω, R4 = 5Ω

and R5 = 2Ω. Calculate the current I3 in the resistor R2 of the circuit shown in the figure

below by using the Thevenin’s theorem.

R3

R1 R4−

R2

R5

+

V3−+

V1

I3

I1

I4

V2+ −

V4

+

−

Ia

I2

Ib I5

1.11 Question no 11

Given I = 24A, R1 = 3Ω, R2 = 6Ω, R3 = 9Ω and R4 = 10Ω. Calculate the current IR4 in

the resistor R4 of the circuit shown in the figure below by using the Norton’s theorem.

1-6

1. Circuit Theorems

R2

R1 R3 R4

a

b

I

1.12 Question no 12

Given V = 30V , I = 10A, R1 = 3Ω, R2 = 5Ω and R3 = 1Ω. Obtain the Norton equivalent

circuit at ab.

R3

a

b

V

R2R1

I+

−

1.13 Question no 13

Given V = 30V , R1 = 3Ω, R2 = 6Ω, R3 = 4Ω, R4 = 2Ω and R5 = 5Ω. Find the current

in the resistor R5 of the following circuit by Norton’s theorem.

1-7

1. Circuit Theorems

B

AR1 R3

R2

R4R5

V

V12

+

−

+

−

+

−

V1

1.14 Question no 14

Given V1 = 10V , V2 = 20V , I = 5A, R1 = 5Ω, R2 = 1Ω. Find the current in the resistor

R2 of the following circuit by Norton’s theorem.

R2

R1I

V1

+

−

+

−V2

1-8

2 Circuit Theorems

2.1 Question no 1

1. .

a b

c d

R1 R3

R2 R4 R5

V2

I2

− +(b)

a b

c d

R1 R3

R4 R5

V1

I1+

−

(a)

2. Use source transformations or Thevenin theorem to calculate the subcurrents.

* Figure (a): I1 = 0.0636A.

2-1

2. Circuit Theorems

* Figure (b): I2 = −1.744A.

* Final: I = I1 + I2 = 0.0636− 1.744 = −1.68A.

2.2 Question no 2

1. .

R1

R3

R2

R5V2

I1

−+

R4

R1 R3

R2

R5

V1

I2

+

−

R4

(a) (b)

2. .

* we calculate the equivalent resistor and then apply the Ohm’s law for figure (a).

* Figure (a): Re = ((R1//R2 + R3)//R4) + R5

* Figure (a): I1 = V2

Re= 10

2+(3(5+[(6×4)/(6+4)]))/(3+5+[(6×4)/(6+4)])= 2.42A.

* we we use the current divider for figure (b).

* Figure (b): IR1 = 106+(4(5+[(3×2)/(3+2)]))/(4+5+[(3×2)/(3+2)])

= 2.37A.

2-2

2. Circuit Theorems

* Figure (b): IR3 = 2.37 44+5+[(3×2)/(3+2)]

= 0.93A.

* Figure (b): I2 = 0.93 33+2

= 0.56A.

* Final: I = I1 + I2 = 2.42 + 0.56 = 2.98A.

2.3 Question no 3

R2R1I

+ − − +

V1 V2

IR

* source V2 and I are removed

* I′R = 12

4+8= 1.0A

* source V1 and I are removed

* I”R = − 64+8

= −0.5A

* source V1 and V2 are removed

* I′′′R = −6 8

4+8= −4.0A

* Total current

* IR = I′R + I”R + I

′′′R = 1.0− 0.5− 4.0 = −3.5A

2-3

2. Circuit Theorems

2.4 Question no 4

R3R2

R4

I′′′R4

R1

V

+ −

R3

R2I2 R1

I′R4

R4

R2

R3I2

I′′R4

R4R1

(a) (b)

(c)

* Figure (a): sources I1 and V are removed

* IR3 = −5 11+3+[(2×4)/(2+4)]

= −0.9375A

* I′R4

= −0.9375 22+4

= −0.3125A

* Figure (b): sources I2 and V are removed

* IR2 = −10 33+1+[(2×4)/(2+4)]

= −5.625A

* I”R4= −(−5.625 2

2+4) = −(−1.875) = 1.875A

* Figure (c): sources I1 and I2 are removed

2-4

2. Circuit Theorems

* I′′′R4

= 204+([2(1+3)]/(2+1+3))

= 3.75A

* Total current

* IR4 = I′R4

+ I”R4+ I

′′′R4

= −0.3125 + 1.875 + 3.75 = 5.3125A

2.5 Question no 5

R2

I′

R1

V2R3

R4

2I1− +

+

−

I′1

R2

I′′

R1

V1

+ −

R3

R4

2I1− +

I”1

I′

I′′

(a) (b)

* Figure (a): source V1 is removed

* Using mesh analysis and combining R2 and R3.

* (2011

+ 4)I′1 + 2I

′1 − 20

11I′= 0

* (2011

+ 10)I′+ 2I

′1 − 20

11I′1 = 50 + 2I

′1

* I′= 4.575A

* Figure (b): source V2 is removed

2-5

2. Circuit Theorems

* Using mesh analysis.

* 22I2 − 2I” − 20I”1 = 70

* 12I” − 2I2 = 2I”1

* 24I”1 − 20I2 = −2I”1

* I” = 3.425A

* Total current

* I = I′+ I” = 4.575 + 3.425 = 8.0A

2.6 Question no 6

R2

R1

R3V1

+

−

(a)

R2

R1 R3V1

+

−

(b)

* Figure (a): source V2 is removed

* V′=

R1R2R1+R2

R3+R1R2R1+R2

V1 =10×1010+10

10+ 10×1010+10

(6) = 515

(6) = 2V

* Figure (b): source V1 is removed

2-6

2. Circuit Theorems

* V′=

R1R2R1+R2

R3+R1R2R1+R2

V2 =10×1010+10

10+ 10×1010+10

(10) = 515

(10) = 3.33V

* Total voltage

* V = V′+ V ” = 2 + 3.33 = 5.33V

2.7 Question no 7

R1 R3

R2V1

V2

I2

+

−

−+

V3

−

+

I3I1 a

b

1. I1 when R2 = 0.

* sources V2 and V3 are removed

* I′1 = V1

2= 3

2= 1.5A.


* I”1 = −V2

2= −2

2= −1A.


* I′′′1 = 0A.

2-7

2. Circuit Theorems

* I1 = I′1 + I”1 + I

′′′1 = 1.5 + (−1) + 0 = 0.5A.

2. I2 when R2 = 0.


* I′2 = −V1

2= −3

2= −1.5A.


* I”2 = V2

2+ V2

1= 2

2+ 2

1= 3A.


* I′′′2 = V3

1= 1

1= 1A.

* I2 = I′2 + I”2 + I

′′′2 = −1.5 + 3 + 1 = 2.5A.

3. I1 when R2 = 1Ω.


* I′1 = V1

2+R2//1= 3

2+(1)(1)/(1+2)= 1.2A.


* I”2 = V2

R2+1//2= 2

2+(1)(1)/(1+2)= 1.2A.

* current divider: I”1 = − 11+2

I”2 = − 11+2

(1.2) = −0.4A.

2-8

2. Circuit Theorems


* I′′′3 = V3

1+R3//2= 1

1+(1)(2)/(1+2)= 0.6A.

* current divider: I′′′1 = R3

R3+2I′′′3 = 1

1+2(0.6) = 0.2A.

* I1 = I′1 + I”1 + I

′′′1 = 1.2 + (−0.4) + 0.2 = 1A.

2.8 Question no 8

R2

R4 R6V1−

R1

R5

+

a

b

R3

R2

R4 R6

R1

R5

a

b

R3

IR3

IR1

IR4

(A) (B)

A. Figure (A): Thevenin voltage

* Re = R1 + [(R4 + R5)//R3] = R1 + [R3(R4+R5)R3+R4+R5

] = 8 + [8(8+4)8+8+4

] = 12.8Ω

* IR1 = VRe

= 9012.8

= 7.03A.

* VR3 = V −R1IR1 = 90− 8IR1 = 90− 7.03(8) = 33.75V .

* IR3 =VR3

R3= 33.75

8= 4.22A.

* IR4 = IR1 − IR3 = 7.03− 4.22 = 2.81A.

2-9

2. Circuit Theorems

* VR4 = VTH = IR4 ×R4 = 2.81× 8 = 22.48V .

B. Figure (B): Thevenin resistor

* RTH = Rab = 6 + (4+4)84+4+8

= 10Ω.

2.9 Question no 9

R2

R3

R4−

R1

R5

+

V3

− +

V2

− +

V1

I1

+

− VTH

IA IB

(a)

R4

R1

R5

RTH

(b)

R2

a. Figure (a): Thevenin voltage

* mesh analysis for IA: V1 + V2 = (R1 + R2)IA + R4(IA − IB) ⇒ V1 + V2 =

(R1 + R2 + R4)IA −R4IB ⇒ 100 + 50 = 14IA − 2IB.

* mesh analysis for IB: −V3+R4(IB−IA)+R5IB = 0⇒ V3 = −R4IA+(R4+R5)IB

⇒ 50 = −2IA + 5IB.

2-10

2. Circuit Theorems

* Solving IA: IA = 12.88A and I1 = IA = 12.88A.

* VTH = V1 − I1R1 = 100− 10(12.88) = −28.8V .

b. Figure (b): Thevenin resistor

* RTH =R1(R2+

R4R5R4+R5

)

R1+R2+R4R5R4+R5

=10(2+ 2×3

2+3)

10+2+ 2×32+3

= 10(2+1.2)10+2+1.2

= 2.42Ω.

⇒ I2 = VTH

RTH+5= −28.8

2.42+5= −3.87A

2.10 Question no 10

VTH

R1 R4−

R2

R5

+

V3−+

V1

I1

I4

V2+ −

V4

+

−

Ia

I2

Ib I5

R1 R4

R2

R5

RTH

(A)

(B)

−+

A. Figure (A): Thevenin voltage

2-11

2. Circuit Theorems

* VTH + V3 − V1 + V4 = 0 ⇒ VTH = −40V

B. Figure (B): Thevenin resistor

* RTH = 0Ω.

* I3 current

* I3 = VTH

20+RTH= −40

20= −2A

2.11 Question no 11

R2

R1 R3 R4I

R2

R1 R3

(a)

(b)

V1

a. Figure (a): Norton current

2-12

2. Circuit Theorems

* current division: IN = R1

R1+R2I = 3

3+624 = 8A

b. Figure (b): Norton resistor

* RN = R3(R1+R2)R3+R1+R2

= 9(3+6)9+3+6

= 4.5Ω.

* IR4 current

* Req = RNR4

RN+R4= 4.5×10

4.5+10= 3.10Ω

* V1 = ReqIN = 3.10× 8 = 24.83V

* IR4 = V1

R4= 24.83

10= 2.5A

2.12 Question no 12

RN

a

b

IN

* Norton current

* ISC = IN = VR1I = 30

3= 10A

2-13

2. Circuit Theorems

* Norton resistor

* RN = R1(R2+R3)R1+R2+R3

= 3(5+1)3+5+1

= 2Ω.

2.13 Question no 13

B

AR1 R3

R2R4 R5V

V12

+

−

+

−

+

−V1

B

AR1 R3

R2R4

R5V

V12

+

−

+

−

+

−V1

ISC

I0

RN

B

A

(a)

(b) (c)

a. Figure a: Norton current

* V1

R3− V1/2

R4− ISC = 0

* ISC = 0A

2-14

2. Circuit Theorems

* IR5 = 0A

b Figure b: voltage output

* I0 = V0−V1

R3+ V0+V1/2

R4= 3

4V0.

c Figure c: Norton resistor

* RN = V0

I0= 4

3Ω.

2.14 Question no 14

VR2

RNIN R2

* Norton current

* I + ISC − V2−V1

R1= 0

* IN = ISC = −3A

* Norton resistor

2-15

2. Circuit Theorems

* RN = 5Ω.

* Norton circuit

*VR2

R2+

VR2

R1= 3

*VR2

1+

VR2

5= 3

* VR2 = 156

= 2.5V

* IR2 =VR2

R2= 2.1

1= 2.5A

2-16

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ETN2B_Chapt3_Homework1