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## 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

Answer: a [Reason:] The capacitor never allows sudden changes in voltage, it will act as a short circuit at t = 0+. So the current in the circuit at t = 0+ is V/R.

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 )

Answer: b [Reason:] The particular solution of the current equation is zero. So the expression of current in R- C circuit is i=(V/R)exp⁡(-t/RC ).

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

Answer: b [Reason:] In a R-C circuit, when the switch is closed, the response decays with time that is the response V/R decreases with increase in time.

4. The time constant of an R-C circuit is?
a) RC
b) R/C
c) R
d) C

Answer: a [Reason:] The time constant of an R-C circuit is RC and it is denoted by τ and the value of τ in dc response of R-C circuit is RC sec.

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

Answer: d [Reason:] After five time constants, the transient part of the response reaches more than 99 percent of its final value.

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

Answer: b [Reason:] At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A.

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

Answer: d [Reason:] By applying Kirchhoff’s law, we get Differentiating with respect to t, we get 10 di/dt+i/0.1=0 => di/dt+i=0.

8. The current equation in the circuit shown in the question 6 is?
a) i=2(e-2t)A
b) i=2(e2t)A
c) i=2(-e-2t)A
d) i=2(-e2t)A

Answer: a [Reason:] At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A. The current equation is i=2(e-2t)A.

9. The expression of voltage across resistor in the circuit shown in the question 6 is?
a) VR =20(et)V
b) VR =20(-e-t)V
c) VR =20(-et)V
d) VR =20(e-t)V

Answer: d [Reason:] The expression of voltage across resistor in the circuit is VR = 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) VC =60(1-e-t )V
b) VC =60(1+et )V
c) VC =60(1-et )V
d) VC =60(1+e-t )V

Answer: a [Reason:] The expression of voltage across capacitor in the circuit VC = 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))

Answer: d [Reason:] The expression of current in R- L circuit is i = (V/R)-(V/R)exp⁡((R/L)t). On solving we get i = (V/R)(1-exp⁡((R/L)t) ).

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

Answer: c [Reason:] The steady state part in the expression of current in the R-L circuit is steady state part = V/R. When the switch S is closed, the response reaches a steady state value after a time interval.

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

Answer: b [Reason:] The expression of current in the R-L circuit has the transient part as (V/R)(-exp⁡((R/L)t) ). The transition period is defined as the time taken for the current to reach its final or steady state value from its initial value.

4. The value of the time constant in the R-L circuit is?
a) L/R
b) R/L
c) R
d) L

Answer: a [Reason:] The time constant of a function (V/R)e-(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

Answer: d [Reason:] After five time constants, the transient part of the response reaches more than 99 percent of its final value.

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

Answer: d [Reason:] Since the inductor never allows sudden changes in currents. At t = 0+ that just after the initial state the current in the circuit is zero.

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

Answer: c [Reason:] Let the i be the current flowing through the circuit. By applying Kirchhoff’s voltage law, we get 15 di/dt+30i=60 => di/dt+2i=4.

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+e2t)A
d) i=2(1+e2t)A

Answer: a [Reason:] At t = 0+ the current in the circuit is zero. Therefore at t = 0+, i = 0 => 0 = c + 2 =>c = -2. Substituting the value of ‘c’ in the current equation, we have i = 2(1-e-2t)A.

9. The expression of voltage across resistor in the circuit shown in the question 6 is?
a) VR =60(1+e2t)V
b) VR =60(1-e-2t)V
c) VR =60(1-e2t)V
d) VR =60(1+e-2t)V

Answer: b [Reason:] Voltage across the resistor VR = 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) VL =60(-e-2t)V
b) VL =60(e2t)V
c) VL =60(e-2t)V
d) VL =60(-e2t)V

Answer: c [Reason:] Voltage across the inductor VL = 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 – (K1 + K2)][D – (K1 – K2)] i = 0. If K2 is positive, then the curve will be?
a) damped
b) over damped
c) under damped
d) critically damped

Answer: b [Reason:] For an R-L-C circuit, we get [D – (K1 + K2)][D – (K1 – K2)] i = 0. If K2 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

Answer: c [Reason:] If the roots of an equation are real and unequal, then the response will be over damped response. Over damped response of a system is defined as the system returns (exponentially decays) to equilibrium without oscillating.

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

Answer: d [Reason:] If the roots of an equation are complex conjugate, then the response will be under damped response. Damping is an influence within or upon an oscillatory system that has the effect of reducing, restricting or preventing its oscillations.

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

Answer: c [Reason:] If the roots of an equation are real and equal, then the response will be critically damped response. For a critically damped system, the system returns to equilibrium as quickly as possible without oscillating.

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.

Answer: a [Reason:] At t = 0, switch S is closed when the 100V source is applied to the circuit and results in the following differential equation.

6. Replacing the differentiation with D1, D2 in the equation obtained from the question 5. Find the values of D1, D2.
a) 200±j979.8
b) -200±j979.8
c) 100±j979.8
d) -100±j979.8

Answer: b [Reason:] Let the roots of the characteristic equation are denoted by D1, D2. So on differentiating the equation obtained in the question 5, we get D1 = -200+j979.8, D2 = -200-j979.8.

7. The expression of current from the circuit shown in the question 5.
a) i=e-200t [c1 cos979.8t+c2 979.8t]A
b) i=e200t [c1 cos979.8t-c2 979.8t]A
c) i=e-200t [c1 cos979.8t-c2 979.8t]A
d) i=e200t [c1 cos979.8t+c2 979.8t]A

Answer: a [Reason:] The expression of current from the circuit will be i = eK1t[c1cosK1t + c2sinK2t]. So, i=e-200t [c1 cos979.8t+c2 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

Answer: d [Reason:] At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

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

Answer: b [Reason:] At t = 0, that is initially the voltage across the inductor is 100V. => V = 100V. So we can write Ldi/dt = 100.

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

Answer: b [Reason:] On solving the values of c1, c2 are obtained as c1 = 0, c2 = 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?

Answer: a [Reason:] The Laplace transform is a powerful analytical technique that is widely used to study the behavior of linear, lumped parameter circuits. L(f(t)) = F (s)

2. Laplace transform changes the ____ domain function to the _____ domain function.
a) time, time
b) time, frequency
c) frequency, time
d) frequency, frequency

Answer: b [Reason:] Laplace transform changes the time domain function f (t) to the frequency domain function F(s). Similarly Laplace transformation converts frequency domain function F(s) to the time domain function f(t).

3. In the bilateral Laplace transform, the lower limit is?
a) 0
b) 1
c) ∞
d) – ∞

Answer: d [Reason:] If the lower limit is 0, then the transform is referred to as one-sided or unilateral Laplace transform. In the two-sided or bilateral Laplace transform, the lower limit is – ∞.

4. The unit step is not defined at t =?
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] If k is 1, the function is defined as unit step function. And the unit step is not defined at t =0.

5. The total period of the function shown in the figure is 4 sec and the amplitude is 10. Find the function f1 (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)].

Answer: d [Reason:] The function shown in the figure is made up of linear segments with break points at 0, 1,3 and 4 seconds. From the graph, f1 (t) = 10t for 0 < t < 1. In terms of unit step function, f1 (t) = 10t [u (t) – u (t – 1)].

6. Find the function f2 (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)].

Answer: b [Reason:] From the graph, f2 (t) = -10t + 20 for 1 < t < 3. In terms of unit step function, f2 (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 f3 (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)].

Answer: a [Reason:] From the graph, f3 (t) = 20t – 40 for 3 < t < 4. In terms of unit step function, f3 (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)].

Answer: c [Reason:] We use the step function to initiate and terminate these linear segments at the proper times. The expression of f (t) is f (t) = 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)].

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

a) 2t
b) 3t
c) 4t
d) 5t

Answer: c [Reason:] The waveform shown in the figure starts at t = 0 and ends at t = 5 sec. The equation for the above waveform is f (t) = 4t.

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)].

Answer: c [Reason:] The waveform shown in the figure starts at t = 0 and ends at t = 5 sec. In terms of unit step function the waveform can be expressed as f (t) = 4t [u (t) – u (t – 5)].

## 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

Answer: b [Reason:] The equation for sine wave A is v(t) = 10sinωt. The value at 90⁰ in this wave is v (t) = 10sin90⁰ = 10V.

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

Answer: a [Reason:] The equation for sine wave A is v(t) = 10sinωt. The equation for sine wave A is v(t) = 8sin(ωt-π/4). ωt = π/2. v (t) = 8sin(π/2-π/4) = 8sin45⁰=8(0.707) = 5.66V.

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

Answer: c [Reason:] The value of the sine wave at some particular instant is called instantaneous value. This value is different at different points along the waveform.

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

Answer: b [Reason:] The maximum value of the wave during positive half cycle or maximum value of the wave during negative cycle is called peak value. Since the values of these two are equal in magnitude, a sine wave is characterized by a single peak value.

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

Answer: c [Reason:] The total area under the complete curve divided by the distance of the curve is called average value. The average value of a sine wave over on e complete cycle is always zero.

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

Answer: d [Reason:] The value from positive to negative peak of the sine wave is called peak to peak value of a sine wave.

7. The RMS value of sine wave is?
a) 0.707Vp
b) 0.607Vp
c) 0.807Vp
d) 0.907Vp

Answer: a [Reason:] The root mean square value of a sine wave is a measure of the heating effect of the wave. The RMS value of sine wave is 0.707Vp.

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

Answer: b [Reason:] Direct current = 20A, sinusoidal alternating current of peak value = 20A. The rms value of the combined wave=√(202+202/2) =24.5A.

9. The peak factor of the sinusoidal waveform is?
a) 4
b) 2
c) 1.414
d) 8