This set of Tricky Network Theory Questions and Answers focuses on “Advanced Problems on Magnetically Coupled Circuits – 2”.

1. In the circuit given below, the resonant frequency is ____________

a) \(\frac{1}{2π\sqrt{3}}\) Hz

b) \(\frac{1}{4π\sqrt{3}}\) Hz

c) \(\frac{1}{4π\sqrt{2}}\) Hz

d) \(\frac{1}{2π\sqrt{6}}\) Hz

View Answer

Explanation: f = \(\frac{1}{2π\sqrt{L_{EQ} C}} \)

Here, L

_{EQ}= 2 + 2 + 2 × 1 = 6

So, C = 1 F

F

_{R}= \(\frac{1}{2π\sqrt{6 × 1}} = \frac{1}{2π\sqrt{6}}\) Hz.

2. A coil is designed for high Q performance at a rated voltage and a specified frequency. If the frequency is made twice the original and the coil is operated at the same rated voltage, then the Q factor will be affected as ____________

a) Q is halved

b) Q remains unchanged

c) Q is doubled

d) Q increases or decreases but magnitude cannot be measured

View Answer

Explanation: ω

_{2}L = 2ω

_{1}L

∴ Q

_{2}= \(\frac{2ω_1 L}{R}\) = 2Q

_{1}

∴ Q is doubled.

3. A coil is designed for high Q performance at a rated voltage and a specified frequency. If the frequency is made twice the original and the coil is operated at the same rated voltage, then the active power P will be affected as ____________

a) P is halved

b) P remains unchanged

c) P is doubled

d) P decreases 4 times

View Answer

Explanation: I

_{1}= \(\frac{V}{\sqrt{R^2+ ω_1^2 L^2}} = \frac{V}{ω_1 L}\)

For a high current coil, ωL >> R

I

_{2}= \(\frac{V_1}{2ω_1 L} = \frac{I_1}{2}\)

∴ P

_{2}= R (\(\frac{I_1}{2}\))

^{2}= \(\frac{P_1}{4}\)

Therefore, P decreases 4 times.

4. In the figure given below, the time constant of the circuit is ______________

a) 2RC

b) 3RC

c) \(\frac{RC}{2}\)

d) \(\frac{2RC}{3}\)

View Answer

Explanation: The simplified circuit is:

Resistance faced by C with the source shorted,

R

_{EQ}= \(\frac{R × 2R}{3R} = \frac{2R}{3}\)

Time constant of the circuit, τ = R

_{EQ}× C

= \(\frac{2R}{3} × C = \frac{2}{3}\) RC.

5. The effective inductance of the circuit across the terminals A, B is _______________

a) 9 H

b) 21 H

c) 11 H

d) 6 H

View Answer

Explanation: Effective inductance across AB terminals

= L

_{1}+ L

_{2}+ L

_{3}– 2M

_{12}– 2M

_{13}+ 2M

_{23}

= 4 + 5 + 6 – 2(1) – 2(3) + 2(2)

= 15 + 4 – 2 – 6 = 11 H.

6. The inductance of a certain moving- iron ammeter is expressed as L = 10 + 3θ – \(\frac{θ^2}{4}\) μH, where θ is the deflection in radian from the zero position. The control spring torque is 25 × 10^{-6} Nm/rad. If the meter is carrying a current of 5 A, the deflection is ____________

a) 2.4

b) 2.0

c) 1.2

d) 1.0

View Answer

Explanation: At equilibrium,

Kθ = \(\frac{1}{2} I^2 \frac{dL}{dθ}\)

(25 × 10

^{-6}) θ = \(\frac{1}{2} I^2 (3 – \frac{θ}{2}) × 10^{-6}\)

∴ 2 θ + \(\frac{θ}{2}\) = 3

Or, θ = 1.2.

7. A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type AC voltmeter connected in parallel. If the meter readings are V_{A} and V_{B} respectively. Then the form factor may be estimated as?

a) \(\frac{V_A}{V_B}\)

b) \(\frac{1.11V_A}{V_B}\)

c) \(\frac{\sqrt{2} V_A}{V_B}\)

d) \(\frac{πV_A}{V_B}\)

View Answer

Explanation: Form factor of the wave = \(\frac{RMS \,value}{Mean \,value}\)

Moving iron instrument will show rms value. Rectifier voltmeter is calibrated to read rms value of sinusoidal voltage that is, with form factor of 1.11.

∴ Mean value of the applied voltage = \(\frac{V_B}{1.11}\)

∴ Form factor = \(\frac{V_A}{V_B/1.11} = \frac{1.11V_A}{V_B}\).

8. A (350 A/7A), 50 Hz current transformer has a primary bar. The secondary has a pure resistance of 1 Ω. It also draws a current of 5 A. The magnetic core requires 350 AT for magnetization. Find the percentage ratio error.

a) 10.56

b) -28.57

c) 11.80

d) -11.80

View Answer

Explanation: I

_{m}= 350/1 =350 A

I

_{p}= \(((nI_s^2)^2 + (I_m^2)^2)^{0.5}\) = 490.05

Or, n = \(\frac{350}{7}\) = 50

∴ R = \(\frac{I_P}{I_S} = \frac{490.05}{7}\) = 70

∴ Percentage ratio error = \(\frac{50-70}{70}\) × 100 = -28.57%.

9. The CT supplies current to the current coil of a wattmeter 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 of a wattmeter, 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.

10. A current of [2 + \(\sqrt{2}\)sin (314t + 30) + 2\(\sqrt{2}\)cos (952t +45)] is measured with a thermocouple type, 5A full scale, class 1 meter. The meter reading would lie in the range?

a) 5 A ± 1 %

b) (2 + 3\(\sqrt{2}\)) A ± 1%

c) 3 A ± 1.7 %

d) 2 A ± 2.5 %

View Answer

Explanation: I = [2 + \(\sqrt{2}\) sin (314t +30°) + 2\(\sqrt{2}\) cos (952t + 45°)]

Thermocouple measure the rms value of current.

I

_{rms}= \(\Big[2^2 + \left(\frac{\sqrt{2}}{\sqrt{2}}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{2}}\right)^2\Big]^{1/2} = \sqrt{9}\) = 3 A ± 1.7%.

11. The average power absorbed by an impedance Z = 30 – j 70 Ω when a voltage V = 120∠0° is applied is _____________

a) 35

b) 37.24

c) 45

d) 50.25

View Answer

Explanation: The current through the impedance is given by,

I = \(\frac{V}{Z} = \frac{120∠0°}{30-j70}\)

= \(\frac{120∠0°}{76.16∠-66.8°}\)

= 1.576∠66.8° A

The average power is, P = 0.5V

_{m}I

_{m}cos (θ

_{v}– θ

_{i})

= 0.5(120) (1.576) cos (0 – 66.8°)

= 37.24 W.

12. A moving iron ammeter produces a full-scale torque of 240 μN-m with a deflection of 120° at a current of 10 A. the rate of change of self-inductance (μH/rad) of the instrument at full scale is?

a) 2.0 μH/rad

b) 4.8 μH/rad

c) 12.0 μH/rad

d) 114.6 μH/rad

View Answer

Explanation: At full scale position, \(\frac{1}{2} I^2 \frac{dL}{dθ}\) = T

_{C}

\(\frac{1}{2} 10^2 \frac{dL}{dθ}\) = 240 × 10

^{-6}

∴ \(\frac{dL}{dθ}\) = 4.8 μH/rad.

13. The relation between the Q factor of a coil measured by the Q Meter and the actual Q of the coil is _________

a) Equal to

b) Same but somewhat lesser than

c) Same but somewhat higher than

d) Not equal to

View Answer

Explanation: The Q factor measured by the Q meter cannot be exactly equal to the actual Q of the coil because of the presence of errors. Also, it is not practically possible for the value to be higher than the actual one. But the value is somewhat lesser and almost equal to the actual value.

14. Consider a circuit consisting of two capacitors C_{1} and C_{2}. Let R be the resistance and L be the inductance which are connected in series. Let Q_{1} and Q_{2} be the quality factor for the two capacitors. While measuring the Q value by the Series Connection method, the value of the Q factor is?

a) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)

b) Q = \(\frac{(C_2 – C_1) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\)

c) Q = \(\frac{(C_1 – C_2) Q_1 Q_2}{Q_2 C_2-Q_1 C_1}\)

d) Q = \(\frac{(C_2 – C_1) C_1 C_2}{Q_1 C_1-Q_2 C_2}\)

View Answer

Explanation: ωL = \(\frac{1}{ωC}\) and Q

_{1}= \(\frac{ωL}{R} = \frac{1}{ωC_1 R}\)

X

_{S}= \(\frac{C_1 – C_2}{ωC_1 C_2}\), R

_{S}= \(\frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2}\)

Q

_{X}= \(\frac{X_S}{R_S} = \frac{(C_1 – C_2) Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

15. The meter constant of a single-phase, 230 V induction watt-meter is 600 rev/kW-h. The speed of the meter disc for a current of 15 A at 0.8 power factor lagging will be?

a) 30.3 rpm

b) 25.02 rpm

c) 27.6 rpm

d) 33.1 rpm

View Answer

Explanation: Meter constant = \(\frac{Number \,of \,revolution}{Energy} = \frac{600 × 230 × 15 × 0.8}{1000}\) = 1656

∴ Speed in rpm = \(\frac{1656}{60}\) = 27.6 rpm.

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