## Network Theory MCQ Set 1

1. The current in the R-L circuit at a time t = 0+ is?

a) V/R

b) R/V

c) V

d) R

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2. The expression of current in R- C circuit is?

a) i=(V/R)exp(t/RC )

b) i=(V/R)exp(-t/RC )

c) i=(V/R)-exp(t/RC )

d) i=(V/R)-exp(-t/RC )

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3. In an R-C circuit, when the switch is closed, the response ____________

a) do not vary with time

b) decays with time

c) rises with time

d) first increases and then decreases

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4. The time constant of an R-C circuit is?

a) RC

b) R/C

c) R

d) C

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5. After how many time constants, the transient part reaches more than 99 percent of its final value?

a) 2

b) 3

c) 4

d) 5

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6.A series R-C circuit consists of resistor of 10 and capacitor of 0.1F as shown in the figure. A constant voltage of 20V is applied to the circuit at t = 0. What is the current in the circuit at t = 0?

a) 1

b) 2

c) 3

d) 4

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7. The expression of current obtained from the circuit in terms of differentiation from the circuit shown in the question 6?

a) di/dt+i=1

b) di/dt+i=2

c) di/dt+i=3

d) di/dt+i=0

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8. The current equation in the circuit shown in the question 6 is?

a) i=2(e^{-2t})A

b) i=2(e^{2t})A

c) i=2(-e^{-2t})A

d) i=2(-e^{2t})A

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^{-2t})A.

9. The expression of voltage across resistor in the circuit shown in the question 6 is?

a) V_{R} =20(e^{t})V

b) V_{R} =20(-e^{-t})V

c) V_{R} =20(-e^{t})V

d) V_{R} =20(e^{-t})V

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_{R}= iR =(2(e

^{-t}) )×10=20(e

^{-t})V.

10. Determine the voltage across the capacitor in the circuit shown in the question 6 is?

a) V_{C} =60(1-e^{-t} )V

b) V_{C} =60(1+e^{t} )V

c) V_{C} =60(1-e^{t} )V

d) V_{C} =60(1+e^{-t} )V

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_{C}= V(1-e

^{-t/RC}) =20(1-e

^{-t})V.

## Network Theory MCQ Set 2

1. The expression of current in R- L circuit is?

a) i=(V/R)(1+exp((R/L)t))

b) i=-(V/R)(1-exp((R/L)t))

c) i=-(V/R)(1+exp((R/L)t))

d) i=(V/R)(1-exp((R/L)t))

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2. The steady state part in the expression of current in the R-L circuit is?

a) (V/R)(exp((R/L)t))

b) (V/R)(-exp((R/L)t))

c) V/R

d) R/V

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3. In the expression of current in the R-L circuit the transient part is?

a) R/V

b) (V/R)(-exp((R/L)t))

c) (V/R)(exp((R/L)t))

d) V/R

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4. The value of the time constant in the R-L circuit is?

a) L/R

b) R/L

c) R

d) L

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^{-(R/L)t}is the time at which the exponent of e is unity where e is the base of the natural logarithms. The term L / R is called the time constant and is denoted by ‘τ’.

5. After how many time constants, the transient part reaches more than 99 percent of its final value?

a) 2

b) 3

c) 4

d) 5

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6. A series R-L circuit with R = 30Ω and L = 15H has a constant voltage V = 60V applied at t = 0 as shown in the figure. Determine the current (A) in the circuit at t = 0+.

a) 1

b) 2

c) 3

d) 0

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7. The expression of current obtained from the circuit in terms of differentiation from the circuit shown in the question 6?

a) di/dt+i=4

b) di/dt+2i=0

c) di/dt+2i=4

d) di/dt-2i=4

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8. The expression of current from the circuit shown in the question 6 is?

a) i=2(1-e^{-2t})A

b) i=2(1+e^{-2t})A

c) i=2(1+e^{2t})A

d) i=2(1+e^{2t})A

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^{-2t})A.

9. The expression of voltage across resistor in the circuit shown in the question 6 is?

a) V_{R} =60(1+e^{2t})V

b) V_{R} =60(1-e^{-2t})V

c) V_{R} =60(1-e^{2t})V

d) V_{R} =60(1+e^{-2t})V

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_{R}= iR. On substituting the expression of current we get voltage across resistor = (2(1-e

^{-2t}) )×30=60(1-e

^{-2t})V.

10. Determine the voltage across the inductor in the circuit shown in the question 6 is?

a) V_{L} =60(-e^{-2t})V

b) V_{L} =60(e^{2t})V

c) V_{L} =60(e^{-2t})V

d) V_{L} =60(-e^{2t})V

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_{L}= Ldi/dt. On substituting the expression of current we get voltage across the inductor = 15×(d/dt)(2(1-e

^{-2t})))=60(e

^{-2t})V.

## Network Theory MCQ Set 3

1. For an R-L-C circuit, we get [D – (K_{1} + K_{2})][D – (K_{1} – K_{2})] i = 0. If K_{2} is positive, then the curve will be?

a) damped

b) over damped

c) under damped

d) critically damped

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_{1}+ K

_{2})][D – (K

_{1}– K

_{2})] i = 0. If K

_{2}is positive, then the curve will be over damped response.

2. If the roots of an equation are real and unequal, then the response will be?

a) critically damped

b) under damped

c) over damped

d) damped

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3. If the roots of an equation are complex conjugate, then the response will be?

a) over damped

b) critically damped

c) damped

d) under damped

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4. If the roots of an equation are real and equal, then the response will be?

a) over damped

b) damped

c) critically damped

d) under damped

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5. The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the switch is closed at t = 0. Find the equation obtained from the circuit in terms of current.

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6. Replacing the differentiation with D_{1}, D_{2} in the equation obtained from the question 5. Find the values of D_{1}, D_{2}.

a) 200±j979.8

b) -200±j979.8

c) 100±j979.8

d) -100±j979.8

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_{1}, D

_{2}. So on differentiating the equation obtained in the question 5, we get D

_{1}= -200+j979.8, D

_{2}= -200-j979.8.

7. The expression of current from the circuit shown in the question 5.

a) i=e^{-200t} [c_{1} cos979.8t+c_{2} 979.8t]A

b) i=e^{200t} [c_{1} cos979.8t-c_{2} 979.8t]A

c) i=e^{-200t} [c_{1} cos979.8t-c_{2} 979.8t]A

d) i=e^{200t} [c_{1} cos979.8t+c_{2} 979.8t]A

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^{K1t}[c

_{1}cosK

_{1}t + c

_{2}sinK

_{2}t]. So, i=e

^{-200t}[c

_{1}cos979.8t+c

_{2}979.8t]A.

8. At time t = 0, the value of current in the circuit shown in the question 5.

a) 1

b) 2

c) 3

d) 0

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9. The voltage across the inductor at t = 0 in the circuit shown in the question 5.

a) 50

b) 100

c) 150

d) 200

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10. The current equation obtained from the circuit shown in the question 5.

a) i=e^{-200t} (1.04 sin979.8t)A

b) i=e^{-200t} (2.04 sin979.8t)A

c) i=e^{-200t} (3.04 sin979.8t)A

d) i=e^{-200t} (4.04 sin979.8t)A

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_{1}, c

_{2}are obtained as c

_{1}= 0, c

_{2}= 2.04. So, the current equation is i=e

^{-200t}(2.04 sin979.8t)A.

## Network Theory MCQ Set 4

1. The Laplace transform of a function f (t) is?

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2. Laplace transform changes the ____ domain function to the _____ domain function.

a) time, time

b) time, frequency

c) frequency, time

d) frequency, frequency

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3. In the bilateral Laplace transform, the lower limit is?

a) 0

b) 1

c) ∞

d) – ∞

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4. The unit step is not defined at t =?

a) 0

b) 1

c) 2

d) 3

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5. The total period of the function shown in the figure is 4 sec and the amplitude is 10. Find the function f_{1} (t) from t = 0 to 1 in terms of unit step function.

a) 10t [u (t) – u (t + 1)].

b) 10t [u (t) + u (t – 1)].

c) 10t [u (t) + u (t + 1)].

d) 10t [u (t) – u (t – 1)].

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_{1}(t) = 10t for 0 < t < 1. In terms of unit step function, f

_{1}(t) = 10t [u (t) – u (t – 1)].

6. Find the function f_{2} (t) from the time t = 1 to 3 sec.

a) (-10t+20) [u (t-1) +u (t-3)].

b) (-10t+20) [u (t-1) – u (t-3)].

c) (-10t-20) [u (t-1) + u (t-3)].

d) (-10t-20) [u (t-1) – u (t-3)].

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_{2}(t) = -10t + 20 for 1 < t < 3. In terms of unit step function, f

_{2}(t) = (-10t+20) [u (t-1) – u (t-3)]. This function turn off at t = 1, turn off at t = 3.

7. Find the function f_{3} (t) from the time t = 3 to 4 sec.

a) (20t – 40) [u (t-3) – u (t-4)].

b) (20t + 40) [u (t-3) – u (t-4)].

c) (20t + 40) [u (t-3) + u (t-4)].

d) (20t – 40) [u (t-3) + u (t-4)].

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_{3}(t) = 20t – 40 for 3 < t < 4. In terms of unit step function, f

_{3}(t) = (20t – 40) [u (t-3) – u (t-4)]. This function turn off at t = 3, turn off at t = 4.

8. Find the expression of f (t) in the graph shown in question 5.

a) 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)].

b) 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)].

c) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)].

d) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)].

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9. In the graph shown below, find the expression f (t).

a) 2t

b) 3t

c) 4t

d) 5t

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10. Find the function f (t) in terms of unit step function in the graph shown in question 9.

a) 4t [u (t) – u (t + 5)].

b) 4t [u (t) + u (t + 5)].

c) 4t [u (t) – u (t – 5)].

d) 4t [u (t) + u (t – 5)].

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## Network Theory MCQ Set 5

1. Find the instantaneous value of the sine value at 90⁰ point having an amplitude 10V and time period 360⁰.

a) 5

b) 10

c) 15

d) 20

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2. Find the instantaneous value at 90⁰ point of the sine wave if the wave is shifted by 45⁰.

a) 5.66

b) 6.66

c) 7.66

d) 8.66

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3. The value of the sine wave at some particular instant is called?

a) peak value

b) peak to peak value

c) instantaneous value

d) average value

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4. The maximum value of the wave during positive half cycle or maximum value of the wave during negative cycle is called?

a) instantaneous value

b) peak value

c) peak to peak value

d) average value

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5. The total area under the complete curve divided by the distance of the curve is called?

a) peak to peak value

b) RMS value

c) average value

d) effective value

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6. The value from positive to negative peak of the sine wave is called?

a) effective value

b) average value

c) peak value

d) peak to peak value

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7. The RMS value of sine wave is?

a) 0.707V_{p}

b) 0.607V_{p}

c) 0.807V_{p}

d) 0.907V_{p}

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_{p}.

8. A wire is carrying a direct current of 20A and a sinusoidal alternating current of peak value 20A. Find the rms value of the resultant current.

a) 24

b) 24.5

c) 25

d) 25.5

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^{2}+20

^{2}/2) =24.5A.

9. The peak factor of the sinusoidal waveform is?

a) 4

b) 2

c) 1.414

d) 8

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_{p}/( V

_{rms}) =V

_{p}/(V

_{p}/√2)=√2=1.414.

10. The form factor of the sinusoidal waveform is?

a) 1.11

b) 2.22

c) 3.33

d) 4.44

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_{rms}/V

_{av}=(V

_{p}/√2)/0.637V

_{p}=1.11.