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

Answer: C

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

Answer: C

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

Answer: A

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

Answer: B

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

Answer: A

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

Answer: B

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

Answer: B

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

Answer: A

Clarification: Mutual inductance w_{m} = ω k (sqrt{L_1 L_2})

Or, m = k.2.8 j^{2}

= 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 X_{C} = 10 Ω. Which of the following is possible?

A. The current and voltage are in phase

B. The current leads the voltage by about 6°

C. The current leads the voltage by about 84°

D. The current lags the voltage by about 6°

Answer: B

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

Answer: C

Clarification: We know that the impedance Z, is given by

Z = R^{2} + (X_C^2)

When f is made four times, X_{C} 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

Answer: B

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)

Answer: B

Clarification: Given that, R = 15Ω, X_{C} = (frac{1}{ωC})

= (frac{10^6}{10 X 1}) = 10^{5} Ω

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}) = 10^{2} 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

Answer: C

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

Answer: D

Clarification: We know that, the characteristic impedance Z0 is given by,

Z_{0} = Z_{oc} Z_{sc}

Given that, open circuit impedance, Z_{oc} = 10 Ω and short circuit impedance, Z_{sc} = 20 Ω.

So, Z_{0} = (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

Answer: A

Clarification: Dynamic resistance of the tank circuit, Z_{DY} = (frac{L}{R_LC})

But given that R_{L} = 0

So, Z_{DY} = (frac{L}{0XC}) = ∞

Therefore current through the circuit, I = (frac{V}{∞}) = 0

∴ V_{D} = 0.