250+ TOP MCQs on Advanced Problems on Application of Laplace Transform – 2 and Answers

Network Theory Objective Questions on “Advanced Problems on Application of Laplace Transform – 2”.

1. A capacitor of 110 V, 50 Hz is needed for AC supply. The peak voltage of the capacitor should be ____________
A. 110 V
B. 460 V
C. 220 V
D. 230 V

Clarification: We know that,
Peak voltage rating = 2 (rms voltage rating).
Given that, rms voltage rating = 110 V
So, Peak voltage rating = 110 X 2 V
= 220 V.

2. Given I (t) = 10[1 + sin (- t)]. The RMS value of I(t) is ____________
A. 10
B. 5
C. (sqrt{150})
D. 105

Clarification: Given that, I (t) = 10[1 + sin (- t)]
Now, RMS value = (sqrt{10^2 + frac{10^2}{sqrt{2}}})
= (sqrt{100+50})
= (sqrt{150}).

3. A two branch parallel circuit has a 50 Ω resistance and 10 H inductance in one branch and a 1 μF capacitor in the second branch. It is fed from 220 V ac supply, at resonance, the input impedance of the circuit is _____________
A. 447.2 Ω
B. 500 Ω
C. 235.48 Ω
D. 325.64 Ω

Clarification: We know that,
For a parallel resonance circuit impedance = ([frac{L}{RC}]^{0.5})
Given, R = 50 Ω, L = 10 H and C = 1 μF
= ([frac{10}{50*1}]^{0.5})
So, impedance = 447.2 Ω.

4. The impedance matrices of two, two-port network are given by [3 2; 2 3] and [15 5; 5 25]. The impedance matrix of the resulting two-port network when the two networks are connected in series is ____________
A. [3 5; 2 25]
B. [18 7; 7 28]
C. [15 2; 5 3]
D. Indeterminate

Clarification: Given that two impedance matrices are [3 2; 2 3] and [15 5; 5 25]. Here, the resulting impedance is the sum of the two given impedances.
So resulting impedance = [18 7; 7 28].

5. The electrical energy required to heat a bucket of water to a certain temperature is 10 kWh. If heat losses, are 10%, the energy input is ____________
A. 2.67 kWh
B. 3 kWh
C. 2.5 kWh
D. 3.5 kWh

Clarification: Given that, 10% of input is lost.
So, 0.90 Input = 10
Or, Input = (frac{10}{0.9}) = 11.11 kWh.

6. The current rating of a cable depends on ___________
A. Length of the cable
B. Diameter of the cable
C. Both length and diameter of the cable
D. Neither length nor diameter of the cable

Clarification: We know that Current rating depends only on the area of cross-section. Since the area of cross section is a function of radius, which in turn is a function of diameter, so the current rating depends only on the Diameter of the cable.

7. In an AC circuit, the maximum and minimum values of power factor can be ___________
A. 2 and 0
B. 1 and 0
C. 0 and -1
D. 1 and -1

Clarification: We know that, power factor is maximum and equal to 1 for a purely resistive load. Power factor is minimum and equal to zero for a purely reactive load.
Hence, in an AC circuit, the maximum and minimum values of power factor are 1 and 0 respectively.

8. In the circuit given below, the series circuit shown in figure, for series resonance, the value of the coupling coefficient, K will be?

A. 0.25
B. 0.5
C. 0.1
D. 1

Clarification: Mutual inductance wm = ω k (sqrt{L_1 L_2})
Or, m = k.2.8 j2
= 4k.j
At resonance, Z = 20 – j10 + j2 + j6 + 2.4kj – 2j + 8kj = 0
Or, -2 = -8k
Or, k = (frac{1}{4}) = 0.25.

9. In an RC series circuit R = 100 Ω and XC = 10 Ω. Which of the following is possible?
A. The current and voltage are in phase
D. The current lags the voltage by about 6°

Clarification: In RC circuit the current leads the voltage.
Or, θ = tan-1 (frac{10}{100})
This is nearly equal to 6°
Hence, the current lags the voltage by about 6°.

10. The impedance of an RC series circuit is 12 Ω at f = 50 Hz. At f = 200 Hz, the impedance will be?
A. More than 12
B. Less than 3
C. More than 3 Ω but less than 12 Ω
D. More than 12 Ω but less than 24 Ω

Clarification: We know that the impedance Z, is given by
Z = R2 + (X_C^2)
When f is made four times, XC becomes one-fourth but R remains the same.
So, the impedance is more than 3 Ω but less than 12 Ω.

11. A 3 phase balanced supply feeds a 3-phase unbalanced load. Power supplied to the load can be measured by ___________
A. 2 Wattmeter and 1 Wattmeter
B. 2 Wattmeter and 3 Wattmeter
C. 2 Wattmeter and 2 Wattmeter
D. Only 3 Wattmeter

Clarification: We know that, a minimum of 2 wattmeters is required to measure a 3-phase power. Also, power can be measured by using only one wattmeter in each phase. Hence, power supplied to the load can be measured by 2 Wattmeter and 3 Wattmeter methods.

12. A series RC circuit has R = 15 Ω and C = 1 μF. The current in the circuit is 5 sin 10t. The applied voltage is _____________
A. 218200 cos (10t – 89.99°)
B. 218200 sin (10t – 89.99°)
C. 218200 sin (10t)
D. 218200 cos (10t)

Clarification: Given that, R = 15Ω, XC = (frac{1}{ωC})
= (frac{10^6}{10 X 1}) = 105
Now, θ = tan-1 (frac{X_C}{R}) = 89.99°
We know that, impedance Z is given by,
Z = (sqrt{R^2 + X_C^2}) = 102 kΩ.
Hence, V = 100000(2.182).sin(10 t – 89.99°)
Or, V = 218200 sin (10t – 89.99°).

13. A capacitor stores 0.15C at 5 V. Its capacitance is ____________
A. 0.75 F
B. 0.75 μF
C. 0.03 F
D. 0.03 μF

Clarification: We know that, Q = CV
Given, V = 5 V, Q = 0.15 C
Hence, 0.15 = C(5)
Or, C = 0.03 F.

14. For a transmission line, open circuit and short circuit impedances are 10 Ω and 20 Ω. Then characteristic impedance is ____________
A. 100 Ω
B. 50 Ω
C. 25 Ω
D. 200 Ω

Clarification: We know that, the characteristic impedance Z0 is given by,
Z0 = Zoc Zsc
Given that, open circuit impedance, Zoc = 10 Ω and short circuit impedance, Zsc = 20 Ω.
So, Z0 = (10.20) Ω
= 200 Ω.

15. A circuit excited by voltage V has a resistance R which is in series with an inductor and a capacitor connected in parallel. The voltage across the resistor at the resonant frequency is ___________
A. 0
B. (frac{V}{2})
C. (frac{V}{3})
D. V