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## 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) ip = V/√(R2+(1/ωC+ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC+ωL)/R))
b) ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
c) ip = V/√(R2+(1/ωC+ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
d) ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC+ωL)/R))

Answer: b [Reason:] The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is ip = V/√(R2+(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) ic = c1 e(K1+K2)t + c1 e(K1-K2)t
b) ic = c1 e(K1-K2)t + c1 e(K1-K2)t
c) ic = c1 e(K1+K2)t + c1 e(K2-K1)t
d) ic = c1 e(K1+K2)t +c1e(K1+K2)t

Answer: a [Reason:] From the R-L circuit, we get the characteristic equation as (D2+R/L D+1/LC)=0. The complementary function of the solution i is ic = c1 e(K1+K2)t + c1 e(K1-K2)t.

3. The complete solution of the current in the sinusoidal response of R-L-C circuit is?
a) i = c1 e(K1+K2)t + c1 e(K1-K2)t – V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
b) i = c1 e(K1+K2)t + c1 e(K1-K2)t – V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ-tan-1)⁡((1/ωC-ωL)/R))
c) i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
d) i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ-tan-1)⁡((1/ωC-ωL)/R))

Answer: c [Reason:] The complete solution for the current becomes i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(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

Answer: d [Reason:] By applying Kirchhoff’s voltage law to the circuit, On differentiating the above equation and on solving, we get roots of the characteristic equation as -37.5±j1290.

5. Find the complementary current from the information provided in the question 4.
a) ic = e-37.5t(c1cos1290t + c2sin1290t)
b) ic = e-37.5t(c1cos1290t – c2sin1290t)
c) ic = e37.5t(c1cos1290t – c2sin1290t)
d) ic = e37.5t(c1cos1290t + c2sin1290t)

Answer: a [Reason:] The roots of the charactesistic equation are D1 = -37.5+j1290 and D2 = -37.5-j1290. The complementary current obtained is ic = e-37.5t(c1cos1290t + c2sin1290t).

6. The particular solution from the information provided in the question 4.
a) ip = 0.6cos(500t + π/4 + 88.5⁰)
b) ip = 0.6cos(500t + π/4 + 89.5⁰)
c) ip = 0.7cos(500t + π/4 + 89.5⁰)
d) ip = 0.7cos(500t + π/4 + 88.5⁰)

Answer: d [Reason:] Particular solution is ip = V/√(R2+(1/ωC-ωL)2 ) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R)). ip = 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(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)
b) i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)
c) i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)
d) i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)

Answer: a [Reason:] The complete solution is the sum of the complementary function and the particular integral. So i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

8. The value of the c1 obtained in the complete solution of question 7.
a) -0.5
b) 0.5
c) 0.6
d) -0.6

Answer: b [Reason:] At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, c1 = -0.71cos (133.5⁰) = 0.49.

9. The value of the c2 obtained in the complete solution of question 7.
a) 2.3
b) -2.3
c) 1.3
d) -1.3

Answer: c [Reason:] Differentiating the current equation, we have di/dt = e-37.5t (-1290c1sin1290t + 1290c2cos1290t) – 37.5e-37.5t(c1cos1290t+c2sin1290t) – 0.71x500sin(500t+45o+88.5o). At t = 0, di/dt = 1414. On solving, we get c2 = 1.31.

10. The complete solution of current obtained by substituting the values of c1 and c2 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)

Answer: d [Reason:] The complete solution is the sum of the complementary function and the particular integral. So i = e-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰).

## Network Theory MCQ Set 2

1. If a resistor ZR is connected between R and N, ZBR between R and B, ZRY between R and Y and ZBY between B and Y form a delta connection, then after transformation to star, the impedance at R is?
a) (ZBRZBY)/(ZRY+ZBY+ZBR)
b) (ZRYZBR)/(ZRY+ZBY+ZBR)
c) (ZRYZBY)/(ZRY+ZBY+ZBR)
d) (ZRY)/(ZRY+ZBY+ZBR)

Answer: b [Reason:] After transformation to star, the impedance at R is (ZRYZBR)/(ZRY+ZBY+ZBR).

2. In the question above the impedance at Y is?
a) (ZRY)/(ZRY+ZBY+ZBR)
b) (ZBY)/(ZRY+ZBY+ZBR)
c) (ZRYZBY)/(ZRY+ZBY+ZBR)
d) (ZRYZBR)/(ZRY+ZBY+ZBR)

Answer: c [Reason:] After transformation to star, the impedance at Y is (ZRYZBY)/(ZRY+ZBY+ZBR).

3. In the question above, the impedance at B is?
a) (ZBRZBY)/(ZRY+ZBY+ZBR)
b) (ZRYZBY)/(ZRY+ZBY+ZBR)
c) (ZBY)/(ZRY+ZBY+ZBR)
d) (ZBR)/(ZRY+ZBY+ZBR)

Answer: a [Reason:] After transformation to star, the impedance at Y is (ZBRZBY)/(ZRY+ZBY+ZBR).

4. If the resistors of star connected system are ZR, ZY, ZB then the impedance ZRY in delta connected system will be?
a) (ZRZY+ ZYZB+ ZBZR)/ZB
b) (ZRZY+ ZYZB+ ZBZR)/ZY
c) (ZRZY+ ZYZB+ ZBZR)/ZR
d) (ZRZY+ ZYZB+ ZBZR)/(ZR+ZY )

Answer: a [Reason:] After transformation to delta, the impedance ZRY in delta connected system will be (ZRZY+ ZYZB+ ZBZR)/ZB.

5. If the resistors of star connected system are ZR, ZY, ZB then the impedance ZBY in delta connected system will be?
a) (ZRZY+ ZYZB+ ZBZR)/(ZB+ZY )
b) (ZRZY+ ZYZB+ ZBZR)/ZB
c) (ZRZY+ ZYZB+ ZBZR)/ZY
d) (ZRZY+ ZYZB+ ZBZR)/ZR

Answer: d [Reason:] After transformation to delta, the impedance ZBY in delta connected system will be (Z_R (ZRZY+ ZYZB+ ZBZR)/ZR.

6. If the resistors of star connected system are ZR, ZY, ZB then the impedance ZBR in delta connected system will be?
a) (ZRZY+ ZYZB+ ZBZR)/ZY
b) (ZRZY+ ZYZB+ ZBZR)/R
c) (ZRZY+ ZYZB+ ZBZR)/ZB
d) (ZRZY+ ZYZB+ ZBZR)/(ZB+ZR )

Answer: a [Reason:] After transformation to delta, the impedance ZBR in delta connected system will be (Z_R (ZRZY+ ZYZB+ ZBZR)/ZY.

7. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are ZR = (2+j3) Ω, ZY = (1-j2) Ω, ZB = (3+j4) Ω. Find ZRY.
a) (5.22-j0.82) Ω
b) (-3.02+j8) Ω
c) (3.8-j0.38) Ω
d) (-5.22+j0.82) Ω

Answer: c [Reason:] ZRY = (ZRZY+ ZYZB+ ZBZR)/ZB = (3.8-j0.38) Ω.

8. Find ZBY 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

Answer: b [Reason:] ZBY = (ZRZY+ ZYZB+ ZBZR)/ZR = (5.22-j0.82)Ω.

9. Find ZBR.
a) (5.22-j0.82) Ω
b) (-3.02+j8) Ω
c) (-5.22+j0.82) Ω
d) (3.8-j0.38) Ω

Answer: b [Reason:] ZBR = (ZRZY+ ZYZB+ ZBZR)/ZY = (-3.02+j8) Ω.

10. If a star connected system has equal impedances Z1, then after converting into delta connected system having equal impedances Z2, then?
a) Z2 = Z1
b) Z2 = 2Z1
c) Z2 = 3Z1
d) Z2 = 4Z1

Answer: c [Reason:] If a star connected system has equal impedances Z1, then after converting into delta connected system having equal impedances Z2, then Z2 = 3Z1.

## Network Theory MCQ Set 3

1.Consider the circuit shown below. Find the current I1 (A).

a) 1
b) 1.33
c) 1.66
d) 2

Answer: b [Reason:] Applying Super mesh analysis, the equations will be I2-I1=2 -10+2I1+I2+4=0. On solving the above equations, I1=1.33A.

2. Find the current I2 (A) in the circuit shown in the question 1.
a) 1.33
b) 2.33
c) 3.33
d) 4.33

Answer: c [Reason:] Applying Super mesh analysis, the equations will be I2-I1=2 -10+2I1+I2+4=0. On solving the above equations, I2=3.33A.

3. Consider the circuit shown in the figure. Find the current I1 (A).

a) -1
b) -2
c) -3
d) -4

Answer: c [Reason:] Applying Super mesh analysis, the equations will be I1+I1+10+I2+I2=0. I1+I2=-5. I2-I1=1. On solving, I1=-3A.

4. Find the current I2 (A) in the figure shown in the question 3.
a) -2
b) -1
c) 2
d) 1

Answer: a [Reason:] Applying Super mesh analysis, the equations will be I1+I1+10+I2+I2=0. I1+I2=-5. I2-I1=1. On solving, I2=-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

Answer: a [Reason:] I3-I2=2. As I2=-2A, I3=0A. Th term power is the product of voltage and current. So, power supplied by source= 10×0=0W.

6. Find the current i1 in the circuit shown below.

a) 8
b) 9
c) 10
d) 11

Answer: c [Reason:] The current in the first loop is equal to 10A. So the current i1 in the circuit is i1 = 10A.

7. Find the current i2 in the circuit shown in the question 6.
a) 6.27
b) 7.27
c) 8.27
d) 9.27

Answer: b [Reason:] For 2nd loop, 10 + 2(i2-i3) + 3(i2-i1) =0. For 3rd loop, i3 + 2(i3-i2)=10. As i1=10A, On solving above equations, we get i2=7.27A.

8. Find the current i3 in the circuit shown in the question 6.
a) 8.18
b) 9.18
c) 10.18
d) 8.8

Answer: a [Reason:] For 2nd loop, 10 +2(i2-i3) +3(i2-i1) =0. For 3rd loop, i3 +2(i3-i2)=10. As i1=10A, On solving above equations, we get i3=8.18A.

9. Find the current I1 in the circuit shown below.

a) 8
b) -8
c) 9
d) -9

Answer: b [Reason:] Applying Super Mesh analysis, (10+5)I1 – 10(I2) – 5(I3) =50. 2(I2) + I3 + 5(I3-I1) + 10(I2-I1) =0. I2 – I3 = 2. On solving above equations, we get I1=-8A.

10. Find the current I2 in the circuit shown in the question 9.
a) 5.3
b) -5.3
c) 7.3
d) -7.3

Answer: d [Reason:] Applying Super Mesh analysis, (10+5)I1-10(I2)-5(I3) =50. 2(I2) + I3 + 5(I3-I1) + 10(I2-I1) =0. I2 – I3 =2. On solving above equations, we get I2=-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

Answer: b [Reason:] Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V1-V3)/3+3+(V2-V3)/1-6+V2/5=0. At node 3, (V3-V1)/3+(V3-V2)/1+V3/2=0. Also V1-V2=10. On solving above equations, we get V1= 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

Answer: b [Reason:] Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V1-V3)/3+3+(V2-V3)/1-6+V2/5=0. At node 3, (V3-V1)/3+(V3-V2)/1+V3/2=0. Also V1-V2=10. On solving above equations, we get V2 = 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

Answer: a [Reason:] Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V1-V3)/3+3+(V2-V3)/1-6+V2/5=0. At node 3, (V3-V1)/3+(V3-V2)/1+V3/2=0. Also V1-V2=10. On solving above equations, we get V3 = 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

Answer: c [Reason:] The term power is defined as the product of voltage and current and the power delivered by the source (6A) = V2x6 = 3.72×6 =22.32W.

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

a) 18
b) 19
c) 20
d) 21

Answer: b [Reason:] The equation at node 1 is 10= V1/3+(V1-V2)/2. According to super Node analysis, (V1-V2)/2=V2/1+(V3-10)/5+V3/2V2-V3=20. On solving, we get, V1=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

Answer: a [Reason:] The equation at node 1 is 10= V1/3+(V1-V2)/2 According to super Node analysis, (V1-V2)/2=V2/1+(V3-10)/5+V3/2V2-V3=20. On solving, we get, V2=11.5V.

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

a) 18
b) 20
c) 22
d) 24

Answer: a [Reason:] At node 1, (V1-40-V3)/4+(V1-V2)/6-3-5=0. Applying Super Node Analysis at node 2 and 3, (V2-V1)/6+5+V2/3+V3/5+(V3+40-V1)/4=0. Also V3-V2=20. On solving above equations, V3 = 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

Answer: b [Reason:] The current through 5Ω resistor = V3/5=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

Answer: c [Reason:] The value of the voltage (V) in the equivalent voltage source of the current source the voltage across the terminals A and B is (6)( 5) = 30V.

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

Answer: b [Reason:] The value of the current (A) in the equivalent current source of the voltage source the short circuit current at the terminals A and B is I=60/30=2A.

## 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

Answer: b [Reason:] In Superposition theorem, while considering a source, all other voltage sources are short circuited. This theorem is valid for linear systems.

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

Answer: c [Reason:] In Superposition theorem, while considering a source, all other current sources are open circuited. Superposition theorem is not valid for power responses. It is applicable only for computing voltage and current responses.

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

a) 4
b) 5
c) 6
d) 7

Answer: b [Reason:] Considering the voltage source 20V, 5A current source is open circuited. Now current through 3Ω resistor is 20/(5+4)=2.22A. Now considering the current source 5A, 20V voltage source is short circuited. No current through 3Ω resistor is 5× 5/(4+5)=2.78A. Now finally the current through 3Ω resistor is 2.22 + 2.78 = 5A.

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

Answer: b [Reason:] Short circuiting 20V source, open circuiting 2A source, Voltage at node A is (V-10)/10+V/20+V/7=0 => V=3.41V. => The voltage across 2Ω resistor is V/(7 )×2=0.97V≅1V.

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

Answer: a [Reason:] Short circuiting 10V source, open circuiting 2A source, The voltage at node A is (V-20)/7+V/20+V/10=0 => V = 9.76V. Now the voltage across 2Ω resistor is (V-20)/7×2=-2.92V.

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

Answer: d [Reason:] Short circuiting both 10V, 20V sources, The current through 2Ω resistor is 2× 5/(5+8.67)=0.73A. The voltage across 2Ω resistor is -0.73×2 = -1.46V.

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

Answer: b [Reason:] The algebraic sum of all the voltages obtained by considering individual sources is the voltage across 2Ω resistor. V = 0.97-2.92-1.46 = -3.41V.

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

Answer: b [Reason:] The voltage at node A in the figure is (V-20)/20+(V-10)/10+V/2=0 => V=3.07V. Now short circuiting 10V source, (V-20)/20+V/2+V/10=0 => V=1.5V.

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

Answer: d [Reason:] The voltage at node A is (V-20)/20+(V-10)/10+V/2=0 => V=3.07V. Now short circuiting 20V source, (V-10)/10+V/20+V/2=0 => V=1.5V.

10. Find the voltage across 2Ω resistor in the circuit shown above using Superposition theorem.
a) 1
b) 2
c) 3
d) 4