This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems on Application of Laplace Transform – 1”.

1. The resistance of a 230 V, 100 W lamp is ____________

a) 529 Ω

b) 2300 Ω

c) 5290 Ω

d) 23 Ω

View Answer

Explanation: P = \(\frac{V^2}{R}\)

Or, R = \(\frac{V^2}{P}\)

Or, R = \(\frac{230 X 230}{100}\)

= 529 Ω.

2. A network has two branches in parallel. One branch contains impedance Z_{a} and the other branch has impedance Z_{b}. If it is fed from an AC voltage V of frequency f, the current through Z_{a} depends, on which of the following?

a) V, Z_{a}, Z_{b}

b) V, Z_{a}

c) Z_{a}, Z_{b}

d) V, Z_{a}, f

View Answer

Explanation: We know that, in parallel branches, the current through any branch depends only on V (voltage), Z (impedance) and f (frequency) only.

So, the current through Z

_{a}depends on V, Z

_{a}, and f.

3. Two coils having self-inductances of 1 mH and 2 mH are mutually coupled. The maximum possible mutual inductance is ___________

a) 1.414 mH

b) 2 mH

c) 1 mH

d) 5.5 mH

View Answer

Explanation: We know that, maximum mutual inductance = L

_{1}L

_{2}

Given that, L

_{1}= 1 mH and L

_{2}= 2 mH

So, maximum mutual inductance = (1 X 2)

^{0.5}= \(\sqrt{2}\)

= 1.414 mH.

4. A constant k band pass filter has a pass band from 100 to 400 Hz. The resonant frequency of series and shunt arms is ____________

a) 300 Hz

b) 250 Hz

c) 200 Hz

d) 150 Hz

View Answer

Explanation: We know that,

Resonant frequency of arms in constant k band pass filter = [(f

_{c1})(f

_{c2})]

^{0.5}

Given that, f

_{c1}= 100 Hz and f

_{c2}= 400 Hz

So, Resonant frequency = \(\sqrt{100 X 400}\)

= 200 Hz.

5. An RLC series circuit has Q = 100 and ω_{0} = 20 rad/sec. The bandwidth is ____________

a) 0.2 rad/sec

b) 2 rad/sec

c) 20 rad/sec

d) 2000 rad/sec

View Answer

Explanation: We know that,

Bandwidth = \(\frac{ω_0}{Q}\)

Given that, ω

_{0}= 20 rad/sec and Q = 100

So, Bandwidth = \(\frac{20}{100}\)

= 0.2 rad/sec.

6. In an unloaded transformer, the flux limiting the primary is 10 mWb and secondary is 30 mWb. The coefficient of coupling is ____________

a) 1

b) 0.1

c) 0.33

d) 0.67

View Answer

Explanation: Φ

_{11}= (1 – k) Φ

_{1}

Or, φ

_{11}= φ

_{1}– φ

_{2}

Or, φ

_{11}= 30 – 10 = 20 mWb

Now, 20 mWb = (1 – k) 30m

Or, 0.67 = 1 – k

Or, k = 0.33.

7. Poles and zeros of a driving point function of a network are simple and alternate on jω axis. The network consists of ___________

a) R and C

b) L and C

c) R and L

d) R, L and C

View Answer

Explanation: In network having only L and C, poles and zeros of driving point function are simple and alternate on jω axis.

So, the network consists of L and C.

8. In a two terminals network the open circuit voltage measured at the given terminals is 110 V and short circuit currents at the same terminals 10 A. If a load of 50 Ω resistance is connected at the terminals, load current is ___________

a) 1.8 A

b) 1.25 A

c) 6 A

d) 6.25 A

View Answer

Explanation: R

_{TH}= \(\frac{110}{10}\) = 11 Ω

So, I = \(\frac{110}{50+11}\) = 1.8 A.

9. A coil has resistance R and inductance L. At ω = ∞ the phase angle between voltage and current is _____________

a) 0°

b) 180°

c) 45°

d) 90°

View Answer

Explanation: We know that,

When ω = ∞, X

_{L}= ωL = ∞.

Therefore, θ = tan

^{-1}\(\frac{ωL}{R}\) = 90°.

10. An RLC series circuit has R = 7.07 Ω, L = .707 H and C = 7.07 μF. At Half power frequencies the circuit impedance is ___________

a) 7.07 Ω

b) 10 Ω

c) 14.14 Ω

d) 20 Ω

View Answer

Explanation: We know that,

At half power frequency the circuit impedance is 2 times resistance

Given that, resistance = 7.07 Ω

So, circuit impedance = 2 X 7.07

= 14.14 Ω.

11. Two similar coils have self-inductance of 20 mH each. Coefficient of coupling is 0.4. The mutual inductance M is ______________

a) 2.5 mH

b) 8 mH

c) 7 mH

d) 1 mH

View Answer

Explanation: We know that, the mutual inductance M is given by,

M = k\(\sqrt{L_1 L_2}\)

Given that, k = 0.4, L

_{1}= 20 mH and L

_{2}= 20 mH

So, M = 0.4\(\sqrt{20 X 20}\)

= 8 mH.

12. For any given signal, average power in its 8 harmonic components as 50 mW each and fundamental component also has 50 mV power. Then, average power in the periodic signal is _______________

a) 750

b) 400

c) 100

d) 50

View Answer

Explanation: We know that according to Parseval’s relation, average power is equal to the sum of the average powers in all of its harmonic components.

∴ P

_{avg}= 50×8 = 400.

13. The Current Transformer supplies current to the current coil of a power factor meter, energy meter and, an ammeter. These are connected as?

a) All coils in parallel

b) All coils in series

c) Series-parallel connection with two in each arm

d) Series-parallel connection with one in each arm

View Answer

Explanation: Since the CT supplies the current to the current coil, therefore the coils are connected in series so that the current remains the same. If they were connected in parallel then the voltages would have been same but the currents would not be same and thus efficiency would decrease.

14. A CRO probe has an impedance of 500 kΩ in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in the figure. The measured voltage will be?

a) 3.53 V

b) 3.47 V

c) 5.54 V

d) 7.00 V

View Answer

Explanation: X

_{C}= \(\frac{1}{jCω} = \frac{-j}{2 × 3.14 × 100 × 10^3 × 10 × 10^{-12}}\)

Applying KCL at node,

\(\frac{V_a-10}{100} + \frac{V_a}{100} + \frac{V_a}{500} + \frac{V_a}{-j159}\)

∴ V

_{a}= 4.37∠-15.95°.

15. M(t) = 2, 0≤t<4;

t^{2}, t≥4;

The Laplace transform of W (t) is ___________

a) \(\frac{2}{s} – e^{-4s} \left(\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s}\right)\)

b) \(\frac{2}{s} + e^{-4s} \left(\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s}\right)\)

c) \(\frac{2}{s} – e^{-4s} \left(\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s}\right)\)

d) \(\frac{2}{s} + e^{-4s} \left(\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s}\right)\)

View Answer

Explanation: M (t) = 2 + u (t) (t

^{2}– 2)

L {2 + u (t) (t

^{2}– 2)} = \(\frac{2}{s}\) + L {u (t) (t

^{2}– 2)}

= \(\frac{2}{s} + e^{-4s}\) L {(t+4)

^{2}-2}

= \(\frac{2}{s} + e^{-4s}\) L {t

^{2}+ 8t + 14}

= \(\frac{2}{s} + e^{-4s} \left(\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s}\right)\).

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