## Network Theory MCQ Set 1

1. The particular current obtained from the solution of i in the sinusoidal response of R-L-C circuit is?

a) i_{p} = V/√(R^{2}+(1/ωC+ωL)^{2} ) cos(ωt+θ+tan^{-1})((1/ωC+ωL)/R))

b) i_{p} = V/√(R^{2}+(1/ωC-ωL)^{2} ) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

c) i_{p} = V/√(R^{2}+(1/ωC+ωL)^{2} ) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

d) i_{p} = V/√(R^{2}+(1/ωC-ωL)^{2} ) cos(ωt+θ+tan^{-1})((1/ωC+ωL)/R))

### View Answer

_{p}= V/√(R

^{2}+(1/ωC-ωL)

^{2}) cos(ωt+θ+tan

^{-1})((1/ωC-ωL)/R)).

2. . In the sinusoidal response of R-L-C circuit, the complementary function of the solution of i is?

a) i_{c} = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t}

b) i_{c} = c_{1} e^{(K1-K2)t} + c_{1} e^{(K1-K2)t}

c) i_{c} = c_{1} e^{(K1+K2)t} + c_{1} e^{(K2-K1)t}

d) i_{c} = c_{1} e^{(K1+K2)t} +c_{1}e^{(K1+K2)t}

### View Answer

^{2}+R/L D+1/LC)=0. The complementary function of the solution i is i

_{c}= c

_{1}e

^{(K1+K2)t}+ c

_{1}e

^{(K1-K2)t}.

3. The complete solution of the current in the sinusoidal response of R-L-C circuit is?

a) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t} – V/√(R^{2}+(1/ωC-ωL)^{2} ) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

b) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t } – V/√(R^{2}+(1/ωC-ωL)^{2} ) cos(ωt+θ-tan^{-1})((1/ωC-ωL)/R))

c) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t} + V/√(R^{2}+(1/ωC-ωL)^{2} ) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

d) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t } + V/√(R^{2}+(1/ωC-ωL)^{2} ) cos(ωt+θ-tan^{-1})((1/ωC-ωL)/R))

### View Answer

_{1}e

^{(K1+K2)t}+ c

_{1}e

^{(K1-K2)t}+ V/√(R

^{2}+(1/ωC-ωL)

^{2}) cos(ωt+θ+tan

^{-1})((1/ωC-ωL)/R)).

4. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.

a) -38.5±j1290

b) 38.5±j1290

c) 37.5±j1290

d) -37.5±j1290

### View Answer

5. Find the complementary current from the information provided in the question 4.

a) i_{c} = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t)

b) i_{c} = e^{-37.5t}(c_{1}cos1290t – c_{2}sin1290t)

c) i_{c} = e^{37.5t}(c_{1}cos1290t – c_{2}sin1290t)

d) i_{c} = e^{37.5t}(c_{1}cos1290t + c_{2}sin1290t)

### View Answer

_{1}= -37.5+j1290 and D

_{2}= -37.5-j1290. The complementary current obtained is i

_{c}= e

^{-37.5t}(c

_{1}cos1290t + c

_{2}sin1290t).

6. The particular solution from the information provided in the question 4.

a) i_{p} = 0.6cos(500t + π/4 + 88.5⁰)

b) i_{p} = 0.6cos(500t + π/4 + 89.5⁰)

c) i_{p} = 0.7cos(500t + π/4 + 89.5⁰)

d) i_{p} = 0.7cos(500t + π/4 + 88.5⁰)

### View Answer

_{p}= V/√(R

^{2}+(1/ωC-ωL)

^{2}) cos(ωt+θ+tan

^{-1})((1/ωC-ωL)/R)). i

_{p}= 0.7cos(500t + π/4 + 88.5⁰).

7. The complete solution of current from the information provided in the question 4.

a) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)

b) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)

c) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)

d) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)

### View Answer

^{-37.5t}(c

_{1}cos1290t + c

_{2}sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

8. The value of the c_{1} obtained in the complete solution of question 7.

a) -0.5

b) 0.5

c) 0.6

d) -0.6

### View Answer

_{1}= -0.71cos (133.5⁰) = 0.49.

9. The value of the c_{2} obtained in the complete solution of question 7.

a) 2.3

b) -2.3

c) 1.3

d) -1.3

### View Answer

^{-37.5t}(-1290c

_{1}sin1290t + 1290c

_{2}cos1290t) – 37.5e

^{-37.5t}(c

_{1}cos1290t+c

_{2}sin1290t) – 0.71x500sin(500t+45

^{o}+88.5

^{o}). At t = 0, di/dt = 1414. On solving, we get c

_{2}= 1.31.

10. The complete solution of current obtained by substituting the values of c_{1} and c_{2} is?

a) i = e^{-37.5t}(0.49cos1290t – 1.3sin1290t) + 0.7cos(500t + 133.5⁰)

b) i = e^{-37.5t}(0.49cos1290t – 1.3sin1290t) – 0.7cos(500t + 133.5⁰)

c) i = e^{-37.5t}(0.49cos1290t + 1.3sin1290t) – 0.7cos(500t + 133.5⁰)

d) i = e^{-37.5t}(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰)

### View Answer

^{-37.5t}(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰).

## Network Theory MCQ Set 2

1. If a resistor Z_{R} is connected between R and N, Z_{BR} between R and B, Z_{RY} between R and Y and Z_{BY} between B and Y form a delta connection, then after transformation to star, the impedance at R is?

a) (Z_{BR}Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

b) (Z_{RY}Z_{BR})/(Z_{RY}+Z_{BY}+Z_{BR})

c) (Z_{RY}Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

d) (Z_{RY})/(Z_{RY}+Z_{BY}+Z_{BR})

### View Answer

_{RY}Z

_{BR})/(Z

_{RY}+Z

_{BY}+Z

_{BR}).

2. In the question above the impedance at Y is?

a) (Z_{RY})/(Z_{RY}+Z_{BY}+Z_{BR})

b) (Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

c) (Z_{RY}Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

d) (Z_{RY}Z_{BR})/(Z_{RY}+Z_{BY}+Z_{BR})

### View Answer

_{RY}Z

_{BY})/(Z

_{RY}+Z

_{BY}+Z

_{BR}).

3. In the question above, the impedance at B is?

a) (Z_{BR}Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

b) (Z_{RY}Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

c) (Z_{BY})/(Z_{RY}+Z_{BY}+Z_{BR})

d) (Z_{BR})/(Z_{RY}+Z_{BY}+Z_{BR})

### View Answer

_{BR}Z

_{BY})/(Z

_{RY}+Z

_{BY}+Z

_{BR}).

4. If the resistors of star connected system are Z_{R}, Z_{Y}, Z_{B} then the impedance Z_{RY} in delta connected system will be?

a) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{B}

b) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{Y}

c) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{R}

d) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/(Z_{R}+Z_{Y} )

### View Answer

_{RY}in delta connected system will be (Z

_{R}Z

_{Y}+ Z

_{Y}Z

_{B}+ Z

_{B}Z

_{R})/Z

_{B}.

5. If the resistors of star connected system are Z_{R}, Z_{Y}, Z_{B} then the impedance Z_{BY} in delta connected system will be?

a) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/(Z_{B}+Z_{Y} )

b) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{B}

c) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{Y}

d) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{R}

### View Answer

_{BY}in delta connected system will be (Z_R (Z

_{R}Z

_{Y}+ Z

_{Y}Z

_{B}+ Z

_{B}Z

_{R})/Z

_{R}.

6. If the resistors of star connected system are Z_{R}, Z_{Y}, Z_{B} then the impedance Z_{BR} in delta connected system will be?

a) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{Y}

b) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/_{R}

c) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/Z_{B}

d) (Z_{R}Z_{Y}+ Z_{Y}Z_{B}+ Z_{B}Z_{R})/(Z_{B}+Z_{R} )

### View Answer

_{BR}in delta connected system will be (Z_R (Z

_{R}Z

_{Y}+ Z

_{Y}Z

_{B}+ Z

_{B}Z

_{R})/Z

_{Y}.

7. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z_{R} = (2+j3) Ω, Z_{Y} = (1-j2) Ω, Z_{B} = (3+j4) Ω. Find Z_{RY}.

a) (5.22-j0.82) Ω

b) (-3.02+j8) Ω

c) (3.8-j0.38) Ω

d) (-5.22+j0.82) Ω

### View Answer

_{RY}= (Z

_{R}Z

_{Y}+ Z

_{Y}Z

_{B}+ Z

_{B}Z

_{R})/Z

_{B}= (3.8-j0.38) Ω.

8. Find Z_{BY} in the question 7.

a) (-5.22+j0.82) Ω

b) (5.22-j0.82) Ω

c) (3.8-j0.38) Ω

d) (-3.02+j8) Ω

### View Answer

_{BY}= (Z

_{R}Z

_{Y}+ Z

_{Y}Z

_{B}+ Z

_{B}Z

_{R})/Z

_{R}= (5.22-j0.82)Ω.

9. Find Z_{BR}.

a) (5.22-j0.82) Ω

b) (-3.02+j8) Ω

c) (-5.22+j0.82) Ω

d) (3.8-j0.38) Ω

### View Answer

_{BR}= (Z

_{R}Z

_{Y}+ Z

_{Y}Z

_{B}+ Z

_{B}Z

_{R})/Z

_{Y}= (-3.02+j8) Ω.

10. If a star connected system has equal impedances Z_{1}, then after converting into delta connected system having equal impedances Z_{2}, then?

a) Z_{2} = Z_{1}

b) Z_{2} = 2Z_{1}

c) Z_{2} = 3Z_{1}

d) Z_{2} = 4Z_{1}

### View Answer

_{1}, then after converting into delta connected system having equal impedances Z

_{2}, then Z

_{2}= 3Z

_{1}.

## Network Theory MCQ Set 3

1.Consider the circuit shown below. Find the current I_{1} (A).

a) 1

b) 1.33

c) 1.66

d) 2

### View Answer

_{2}-I

_{1}=2 -10+2I

_{1}+I

_{2}+4=0. On solving the above equations, I

_{1}=1.33A.

2. Find the current I_{2} (A) in the circuit shown in the question 1.

a) 1.33

b) 2.33

c) 3.33

d) 4.33

### View Answer

_{2}-I

_{1}=2 -10+2I

_{1}+I

_{2}+4=0. On solving the above equations, I

_{2}=3.33A.

3. Consider the circuit shown in the figure. Find the current I_{1} (A).

a) -1

b) -2

c) -3

d) -4

### View Answer

_{1}+I

_{1}+10+I

_{2}+I

_{2}=0. I

_{1}+I

_{2}=-5. I

_{2}-I

_{1}=1. On solving, I

_{1}=-3A.

4. Find the current I_{2} (A) in the figure shown in the question 3.

a) -2

b) -1

c) 2

d) 1

### View Answer

_{1}+I

_{1}+10+I

_{2}+I

_{2}=0. I

_{1}+I

_{2}=-5. I

_{2}-I

_{1}=1. On solving, I

_{2}=-2A.

5. Find the power (W) supplied by the voltage source in the figure shown in the question 3.

a) 0

b) 1

c) 2

d) 3

### View Answer

_{3}-I

_{2}=2. As I

_{2}=-2A, I

_{3}=0A. Th term power is the product of voltage and current. So, power supplied by source= 10×0=0W.

6. Find the current i_{1} in the circuit shown below.

a) 8

b) 9

c) 10

d) 11

### View Answer

_{1}in the circuit is i

_{1}= 10A.

7. Find the current i_{2} in the circuit shown in the question 6.

a) 6.27

b) 7.27

c) 8.27

d) 9.27

### View Answer

_{2}-i

_{3}) + 3(i

_{2}-i

_{1}) =0. For 3rd loop, i

_{3}+ 2(i

_{3}-i

_{2})=10. As i

_{1}=10A, On solving above equations, we get i

_{2}=7.27A.

8. Find the current i_{3} in the circuit shown in the question 6.

a) 8.18

b) 9.18

c) 10.18

d) 8.8

### View Answer

_{2}-i

_{3}) +3(i

_{2}-i

_{1}) =0. For 3rd loop, i

_{3}+2(i

_{3}-i

_{2})=10. As i

_{1}=10A, On solving above equations, we get i

_{3}=8.18A.

9. Find the current I_{1} in the circuit shown below.

a) 8

b) -8

c) 9

d) -9

### View Answer

_{1}– 10(I

_{2}) – 5(I

_{3}) =50. 2(I

_{2}) + I

_{3}+ 5(I

_{3}-I

_{1}) + 10(I

_{2}-I

_{1}) =0. I

_{2}– I

_{3}= 2. On solving above equations, we get I

_{1}=-8A.

10. Find the current I_{2} in the circuit shown in the question 9.

a) 5.3

b) -5.3

c) 7.3

d) -7.3

### View Answer

_{1}-10(I

_{2})-5(I

_{3}) =50. 2(I

_{2}) + I

_{3}+ 5(I

_{3}-I

_{1}) + 10(I

_{2}-I

_{1}) =0. I

_{2}– I

_{3}=2. On solving above equations, we get I

_{2}=-7.3A.

## Network Theory MCQ Set 4

1.Consider the figure shown below. Find the voltage (V) at node 1.

a) 13

b) 14

c) 15

d) 16

### View Answer

_{1}-V

_{3})/3+3+(V

_{2}-V

_{3})/1-6+V

_{2}/5=0. At node 3, (V

_{3}-V

_{1})/3+(V

_{3}-V

_{2})/1+V

_{3}/2=0. Also V

_{1}-V

_{2}=10. On solving above equations, we get V

_{1}= 13.72V ≈ 14V.

2. In the figure shown in the question 1 find the voltage (V) at node 2.

a) 3

b) 4

c) 5

d) 6

### View Answer

_{1}-V

_{3})/3+3+(V

_{2}-V

_{3})/1-6+V

_{2}/5=0. At node 3, (V

_{3}-V

_{1})/3+(V

_{3}-V

_{2})/1+V

_{3}/2=0. Also V

_{1}-V

_{2}=10. On solving above equations, we get V

_{2}= 3.72V ≈ 4V.

3. In the figure shown in the question 1 find the voltage (V) at node 3.

a) 4.5

b) 5.5

c) 6.5

d) 7.5

### View Answer

_{1}-V

_{3})/3+3+(V

_{2}-V

_{3})/1-6+V

_{2}/5=0. At node 3, (V

_{3}-V

_{1})/3+(V

_{3}-V

_{2})/1+V

_{3}/2=0. Also V

_{1}-V

_{2}=10. On solving above equations, we get V

_{3}= 4.5V.

4. In the figure shown in the question 1 find the power (W) delivered by the source 6A.

a) 20.3

b) 21.3

c) 22.3

d) 24.3

### View Answer

_{2}x6 = 3.72×6 =22.32W.

5. Find the voltage (V) at node 1 in the circuit

a) 18

b) 19

c) 20

d) 21

### View Answer

_{1}/3+(V

_{1}-V

_{2})/2. According to super Node analysis, (V

_{1}-V

_{2})/2=V

_{2}/1+(V

_{3}-10)/5+V

_{3}/2V

_{2}-V

_{3}=20. On solving, we get, V

_{1}=19V.

6. Find the voltage (V) at node 2 of the circuit shown in the question 5.

a) 11.5

b) 12

c) 12.5

d) 13

### View Answer

_{1}/3+(V

_{1}-V

_{2})/2 According to super Node analysis, (V

_{1}-V

_{2})/2=V

_{2}/1+(V

_{3}-10)/5+V

_{3}/2V

_{2}-V

_{3}=20. On solving, we get, V

_{2}=11.5V.

7. Find the voltage (V) at node 3 in the figure shown below.

a) 18

b) 20

c) 22

d) 24

### View Answer

_{1}-40-V

_{3})/4+(V

_{1}-V

_{2})/6-3-5=0. Applying Super Node Analysis at node 2 and 3, (V

_{2}-V

_{1})/6+5+V

_{2}/3+V

_{3}/5+(V

_{3}+40-V

_{1})/4=0. Also V

_{3}-V

_{2}=20. On solving above equations, V

_{3}= 18.11V ≈ 18V.

8. Find the power absorbed by 5Ω resistor in the figure shown in the question 7.

a) 60

b) 65.5

c) 70.6

d) 75

### View Answer

_{3}/5=18.11/5=3.62A. The power absorbed by 5Ω resistor = (3.62)

^{2})×5=65.52W.

9. Find the value of the voltage (V) in the equivalent voltage source of the current source shown below.

a) 20

b) 25

c) 30

d) 35

### View Answer

10. Find the value of the current (A) in the equivalent current source of the voltage source shown below.

a) 1

b) 2

c) 3

d) 4

### View Answer

## Network Theory MCQ Set 5

1. In Superposition theorem, while considering a source, all other voltage sources are?

a) open circuited

b) short circuited

c) change its position

d) removed from the circuit

### View Answer

2. In Superposition theorem, while considering a source, all other current sources are?

a) short circuited

b) change its position

c) open circuited

d) removed from the circuit

### View Answer

3. In the circuit shown, find the current through 4Ω resistor using Superposition theorem.

a) 4

b) 5

c) 6

d) 7

### View Answer

4. Consider the circuit shown below. Find the voltage across 2Ω resistor due to the 10V voltage source using Superposition theorem.

a) 0

b) 1

c) 2

d) 3

### View Answer

5. Find the voltage across 2Ω resistor due to 20V source in the figure shown above.

a) -2.92

b) 2.92

c) 1.92

d) -1.92

### View Answer

6. Find the voltage across 2Ω resistor due to 2A source in the figure shown above.

a) -1

b) 1

c) 1.46

d) -1.46

### View Answer

7. In the figure shown above find the voltage across 2Ω resistor due to all source using Superposition theorem.

a) 3.41

b) -3.41

c) 3.14

d) -3.14

### View Answer

8. Find the voltage across 2Ω resistor due to 20V source in the circuit shown below .

a) 1

b) 1.5

c) 2

d) 2.5

### View Answer

9. Find the voltage across 2Ω resistor due to 20V source in the circuit shown above.

a) 0.5

b) 0

c) 1

d) 1.5

### View Answer

10. Find the voltage across 2Ω resistor in the circuit shown above using Superposition theorem.

a) 1

b) 2

c) 3

d) 4