## Network Theory MCQ Set 1

1. To check for the Reciprocity Theorem we consider ______ of response to excitation.

a) ratio

b) addition

c) product

d) subtraction

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2. For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be?

a) different

b) same

c) before source is replaced is greater than after the source is replaced

d) before source is replaced is less than after the source is replaced

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3. The circuit which satisfies Reciprocity Theorem is called?

a) Short circuit

b) Open circuit

c) Linear circuit

d) Non-linear circuit

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4. Find the current through the 2Ω(c-d) resistor in the circuit shown below.

a) 0.143

b) 1.43

c) 14.3

d) 143

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_{t})=20/3.5=5.71Ω. The current drawn by 2Ω resistor = 1.43A.

5. The current drawn by 2Ω resistor (a-b) after the source is replaced in the question 4?

a) 143

b) 14.3

c) 1.43

d) 0.143

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6. The circuit in the question 4 satisfies Reciprocity Theorem.

a) True

b) False

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7. Find the current through 3Ω resistor in the circuit shown below.

a) 1

b) 2

c) 3

d) 4

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8. Find the current through 2Ω resistor after source is replaced in the question 7.

a) 4

b) 3

c) 2

d) 1

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9. The circuit shown in the question 7 satisfies the reciprocity theorem.

a) False

b) True

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10. While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of?

a) voltage to voltage

b) current to current

c) voltage to current

d) None of the above

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## Network Theory MCQ Set 2

1. If we apply a sinusoidal input to RL circuit, the current in the circuit is __________ and the voltage across the elements is _______________

a) square, square

b) square, sinusoid

c) sinusoid, square

d) sinusoid, sinusoid

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2. The circuit shown below consists of a 1kΩ resistor connected in series with a 50mH coil, a 10V rms, 10 KHz signal is applied. Find impedance Z in rectangular form.

a) (1000+j0.05) Ω

b) (100+j0.5) Ω

c) (1000+j3140) Ω

d) (100+j3140) Ω

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_{L}= ωL = 2πfL = (6.28)(10

^{4})(50×10

^{-3}) =3140Ω. In rectangular form, total impedance Z = (1000+j3140) Ω.

3. Find the current I (mA) in the circuit shown above.

a) 3.03

b) 30.3

c) 303

d) 0.303

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_{s}/Z = 10/3295.4=3.03mA.

4. Find the phase angle θ in the circuit shown above.

a) 62.33⁰

b) 72.33⁰

c) 82.33⁰

d) 92.33⁰

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^{-1}(X

_{L}/R). The values of X

_{L}, R are X

_{L}= 3140Ω and R = 1000Ω. On substituting the values in the equation, phase angle θ =tan

^{-1}(3140/1000)=72.33⁰.

5. In the circuit shown above find the voltage across resistance.

a) 0.303

b) 303

c) 3.03

d) 30.3

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^{-3}×1000 = 3.03V.

6. In the circuit shown above find voltage across inductive reactance.

a) 9.5

b) 10

c) 9

d) 10.5

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_{L}. The values of I = 3.03 mA and X

_{L}= 10000Ω. On substituting the values in the equation, the voltage across inductor =3.03×10

^{-3}×1000 = 9.51V.

7. Determine the source voltage if voltage across resistance is 70V and the voltage across inductor is 20V as shown in the figure.

a) 71

b) 72

c) 73

d) 74

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_{s}=√(70

^{2}+20

^{2}) = 72.8≅73V.

8. Find the phase angle in the circuit shown above.

a) 15

b) 16

c) 17

d) 18

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^{-1}(V

_{L}/V

_{R}). The values of V

_{L}= 20V and V

_{R}= 70V. On substituting the values in the equation, phase angle in the circuit =tan

^{-1}(20/70)=15.94

^{o}≅ 16

^{o}.

9. An AC voltage source supplies a 500Hz, 10V rms signal to a 2kΩ resistor in series with a 0.1µF capacitor as shown in figure. Find the total impedance.

a) 3750.6Ω

b) 3760.6Ω

c) 3780.6Ω

d) 3790.6Ω

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_{C}= 1/2πfC =1/(6.28×500×0.1×10

^{-6}) )=3184.7Ω. In rectangular form, X = (2000-j3184.7)Ω. Magnitude = 3760.6Ω.

10. Determine the phase angle in the circuit shown above.

a) 58

b) 68

c) -58

d) -68

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^{-1}(-X

_{C}/R) =tan

^{-1}((-3184.7)/2000)=-57.87

^{o}≅-58

^{o}.

11. Find the current I (mA) in the circuit shown above.

a) 2.66

b) 3.66

c) 4.66

d) 5.66

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_{S}/ Z = 10/3760.6 = 2.66mA

12. Find the voltage across the capacitor in the circuit shown above.

a) 7

b) 7.5

c) 8

d) 8.5

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^{-3}×3184.7 =8.47V.

13. Determine the voltage across the resistor in the circuit shown above.

a) 3

b) 4

c) 5

d) 6

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^{-3}×3184.7 =5.32V.

14. In the circuit shown below determine the total impedance.

a) 161

b) 162

c) 163

d) 164

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^{-6}) =318.5Ω. Reactance across inductor = 6.28×0.5×50=157Ω. In rectangular form, Z = (10+j157-j318.5) Ω = (10-j161.5)Ω. Magnitude=161.8Ω.

15. Find the current in the circuit shown above.

a) 0.1

b) 0.2

c) 0.3

d) 0.4

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_{S}/Z = 50/161.8 = 0.3A.

## Network Theory MCQ Set 3

1. The circuit is said to be in resonance if the current is ____ with the applied voltage.

a) in phase

b) out of phase

c) 45⁰ out of phase

d) 90⁰ out of phase

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2. In a series resonance circuit, series resonance occurs when?

a) X_{L} = 1

b) X_{C} = 1

c) X_{L} = X_{C}

d) X_{L} = -X_{C}

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_{L}= X

_{C}.

3. As X_{L} = X_{C} in a series resonance circuit, the impedance is_________

a) purely capacitive

b) purely inductive

c) purely resistive

d) capacitive and inductive

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_{L}= X

_{C}in a series resonance circuit, the impedance is purely resistive. In a series RLC circuit the current lags behind or leads the applied voltage depending on the values of X

_{L}and X

_{C}.

4. At resonant frequency, the voltage across capacitor is _______ the voltage across inductor.

a) greater than

b) less than

c) greater than or equal to

d) equal to

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5. In series RLC circuit, the voltage across capacitor and inductor are ______ with each other.

a) in phase

b) 180⁰ out of phase

c) 90⁰ out of phase

d) 45⁰ out of phase

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6. The voltage across the LC combination in a series RLC circuit is?

a) 0

b) 1

c) 2

d) 3

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7. The expression of resonant frequency in a series resonant circuit is?

a) 1/(2π√C)

b) 1/( 2π√L)

c) 2π√LC

d) 1/(2π√LC)

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8. For the circuit shown in figure determine the capacitive reactance at resonance.

a) 15

b) 20

c) 25

d) 30

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_{L}= X

_{C}. Given inductive reactance X

_{L}= 25. On substituting in the equation we get X

_{C}= 25Ω.

9. The value of the impedance at resonance in the circuit shown in the question 9 is?

a) 25

b) 50

c) 75

d) 100

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10. Determine the resonant frequency (kHz) for the circuit shown below.

a) 2.25

b) 22.5

c) 225

d) 2250

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_{r}= 1/(2π√LC). Given L = 0.5mH and C = 10uF. On substituting in the equation we get resonant frequency f

_{r}= 1/(2π√(10×10

^{-6})×0.5×10

^{-3})) = 2.25kHz.

## Network Theory MCQ Set 4

1. In determining short circuit impedance parameters, among V_{1}, V_{2}, I_{1}, I_{2}, which of the following are dependent variables?

a) V_{1} and V_{2}

b) I_{1} and I_{2}

c) V_{1} and I_{2}

d) I_{1} and V_{2}

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_{1}, V

_{2}, I

_{1}, I

_{2}; I

_{1}and I

_{2}are dependent variables and V

_{1}, V

_{2}are independent variables i.e., dependent variables depend on independent variables.

2. In determining short circuit impedance parameters, among V_{1}, V_{2}, I_{1}, I_{2}, which of the following are independent variables?

a) I_{1} and V_{2}

b) V_{1} and I_{2}

c) I_{1} and I_{2}

d) V_{1} and V_{2}

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_{1}, V

_{2}, I

_{1}, I

_{2}; V

_{1}and V

_{2}are independent variables and I

_{1}, I

_{2}are dependent variables. Independent variables are the variables that do not depend on any other variable.

3. Which of the following expression is true in case of short circuit parameters?

a) I_{1} = Y_{11} V_{1} + Y_{12} V_{2}

b) I_{1} = Y_{11} I_{1} + Y_{12} V_{2}

c) V_{1} = Y_{11} I_{1} + Y_{12} V_{2}

d) V_{1} = Y_{11} V_{1} + Y_{12} V_{2}

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_{11}, Y

_{12}and voltages V

_{1}, V

_{2}and current is I

_{1}, is I

_{1}= Y

_{11}V

_{1}+ Y

_{12}V

_{2}.

4. Which of the following expression is true in case of short circuit parameters?

a) I_{2} = Y_{21}I_{1} + Y_{22} I_{2}

b) V_{2} = Y_{21}I_{1} + Y_{22} V_{2}

c) I_{2} = Y_{21}V_{1} + Y_{22} V_{2}

d) I_{2} = Y_{21}V_{1} + Y_{22} I_{2}

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_{1}, V

_{2}and current is I

_{2}and short circuit parameters Y

_{11}, Y

_{12}is I

_{2}= Y

_{21}V

_{1}+ Y

_{22}V

_{2}.

5. The parameters Y_{11}, Y_{12}, Y_{21}, Y_{22} are called?

a) Open circuit impedance parameters

b) Short circuit admittance parameters

c) Inverse transmission parameters

d) Transmission parameters

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_{11}, Y

_{12}, Y

_{21}, Y

_{22}are called short circuit admittance parameters also called network functions as they are obtained by short circuiting port 1 or port 2.

6. Find the Y – parameter Y_{11} in the circuit shown below.

a) 2

b) 3/2

c) 1

d) 1/2

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_{1}= (I

_{1}) x 2. We know Y

_{11}= I

_{1}/V

_{1}. From the equation we get I

_{1}/V

_{1}= 2. On substituting we get Y

_{11}= 2 mho.

7. Find the Y – parameter Y_{21} in the circuit shown in question 6.

a) -1/4

b) 1/4

c) 1/2

d) -1/2

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_{2}=I

_{1}× 2/4=I

_{1}/2 and -I

_{2}= V

_{1}/4 and on solving and substituting we get Y

_{21}=I

_{2}/V

_{1}=-1/4 mho.

8. Find the Y – parameter Y_{22} in the circuit shown in question 6.

a) 3/8

b) 5/8

c) 7/8

d) 9/8

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_{eq}= 8/5 Ω. V

_{2}=I

_{2}× 8/5. We know Y

_{22}= I

_{2}/V

_{2}. We got I

_{2}/V

_{1}= 5/8. ON substituting we get Y

_{22}= 5/8 mho.

9. Find the Y – parameter Y_{12} in the circuit shown in question 6.

a) 1/2

b) -1/2

c) -1/4

d) 1/4

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_{1}= 2/5 I

_{2}and I

_{2}= 5 V

_{2}/8. On solving -I

_{1}= 2/5×5/8 V

_{2}= V

_{2}/4. We know Y

_{12}= I

_{1}/V

_{2}. We got I

_{1}/V

_{2}= -1/4. So the value of Y

_{12}will be -1/4 mho.

10. Which of the following equation is true in the circuit shown in question 6?

a) I_{1}=0.5(V_{1})+0.25(V_{2})

b) I_{1}=0.25(V_{1})+0.625(V_{2})

c) I_{1}=-0.25(V_{1})+0.625(V_{2})

d) I_{1}=0.5(V_{1})-0.25(V_{2})

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_{11}= 0.5, Y

_{12}= -0.25, Y

_{21}= -0.25, Y

_{22}= 0.625. So the equations in terms of admittance parameters is I

_{1}=0.5(V

_{1})-0.25(V

_{2}) and I

_{2}=-0.25(V

_{1})+0.625(V

_{2}).

## Network Theory MCQ Set 5

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

a) i_{c} = ce^{-t(R/L)}

b) i_{c} = ce^{t(RL)}

c) i_{c} = ce^{-t(RL)}

d) i_{c} = ce^{t(R/L)}

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_{c}= ce

^{-t(R/L)}.

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

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

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

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

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

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_{p}= V/√(R

^{2}+(ωL)

^{2}) cos(ωt+θ-tan

^{-1}(ωL/R)).

3. The value of ‘c’ in complementary function of ‘i’ is?

a) c = -V/√(R^{2}+(ωL)^{2}) cos(θ+tan^{-1}(ωL/R))

b) c = -V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1}(ωL/R))

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

d) c = V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1}(ωL/R))

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^{2}+(ωL)

^{2}) cos(θ-tan

^{-1}(ωL/R)).

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

a) i = e^{-t(R/L)}[V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]+V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

b) i = e^{-t(R/L)}[-V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]-V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

c) i = e^{-t(R/L)}[V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]-V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

d) i = e^{-t(R/L)}[-V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]+V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

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^{-t(R/L)}[-V/√(R

^{2}+(ωL)

^{2}) cos(θ-tan

^{-1})(ωL/R))]+V/√(R

^{2}+(ωL)

^{2})cos(ωt+θ-tan

^{-1})(ωL/R)).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complementary function of the solution of ‘i’ is?

a) i_{c} = ce^{-100t}

b) i_{c} = ce^{100t}

c) i_{c} = ce^{-200t}

d) i_{c} = ce^{200t}

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^{3}t+π/2) => (D+200)i=1000cos(1000t+π/2). The complementary function is i

_{c}= ce

^{-200t}.

6. The particular integral of the solution of ‘i’ from the information provided in the question 5.

a) i_{p} = 0.98cos(1000t+π/2-78.6^{o})

b) i_{p} = 0.98cos(1000t-π/2-78.6^{o})

c) i_{p} = 0.98cos(1000t-π/2+78.6^{o})

d) i_{p} = 0.98cos(1000t+π/2+78.6^{o})

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_{p}= A cos (ωt + θ) + B sin(ωt + θ). We get i

_{p}= V/√(R

^{2}+(ωL)

^{2}) cos(ωt+θ-tan

^{-1}(ωL/R)) where ω = 1000 rad/sec, V = 100V, θ = π/2, L = 0.1H, R = 20Ω. On substituting, we get i

_{p}= 0.98cos(1000t+π/2-78.6

^{o}).

7. The complete solution of ‘i’ from the information provided in the question 5.

a) i = ce^{-200t} + 0.98cos(1000t-π/2-78.6^{o})

b) i = ce^{-200t} + 0.98cos(1000t+π/2-78.6^{o})

c) i = ce^{-200t} + 0.98cos(1000t+π/2+78.6^{o})

d) i = ce^{-200t} + 0.98cos(1000t-π/2+78.6^{o})

### View Answer

^{-200t}+ 0.98cos(1000t+π/2-78.6

^{o}).

8. The current flowing through the circuit at t = 0 in the circuit shown in the question 5 is?

a) 1

b) 2

c) 3

d) 0

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9. The value of c in the complementary function of ‘i’ in the question 5 is?

a) c = -0.98cos(π/2-78.6^{o})

b) c = -0.98cos(π/2+78.6^{o})

c) c = 0.98cos(π/2+78.6^{o})

d) c = 0.98cos(π/2-78.6^{o})

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^{o}).

10. The complete solution of ‘i’ in the question 5 is?

a) i = [-0.98 cos(π/2-78.6^{o})] exp(-200t)+0.98cos(1000t+π/2-78.6^{o})

b) i = [-0.98 cos(π/2-78.6^{o})] exp(-200t)-0.98cos(1000t+π/2-78.6^{o})

c) i = [0.98 cos(π/2-78.6^{o})] exp(-200t)-0.98cos(1000t+π/2-78.6^{o})

d) i = [0.98 cos(π/2-78.6^{o})] exp(-200t)+0.98cos(1000t+π/2-78.6^{o})

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^{o})] exp(-200t)+0.98cos(1000t+π/2-78.6

^{o}).