Inductance in DC Circuit MCQ [Free PDF] - Objective Question Answer for Inductance in DC Circuit Quiz

91. What happens to the coupling coefficient when the flux linkage of coil 1 and coil 2 increases?

A. Increases
B. Decreases
C. Remains the same
D. Becomes zero

Answer: A

When the flux linkage of coil 1 and coil 2 increases, its mutual inductance increases. The coupling coefficient is directly proportional to the mutual inductance hence as mutual inductance increases, the coupling coefficient increases.

92. What is the SI unit of the coupling coefficient?

A. H
B. H^{− 1}
C. No unit
D. H²

Answer: C

The expression to find mutual inductance is

k = M/√(L1L2) = H/√(H × H) = 1.

Therefore it does not have any unit.

93. Find the coupling coefficient if the Mutual inductance is 20H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H.

A. 5
B. 20
C. 2
D. 8

Answer: A

We know that coupling coefficient K is

k = M/√(L1L2)

Substituting the values from the question, we get k = 5.

94. Find the value of x if the Mutual inductance is x H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H. The coupling coefficient is 5.

A. 10H
B. 20H
C. 16H
D. 15H

Answer: B

we know that:

k = M/√(L1L2)

Substituting the values from the question, we get M = 20H.

95. Find the value of x if the Mutual inductance is 20H, the inductance of coil 1 is xH and the inductance of coil 2 is 8H. The coupling coefficient is 5.

A. 2H
B. 4H
C. 6H
D. 8H

Answer: A

We know that coupling coefficient K is

k = M/√(L1L2)

Substituting the values from the question, we get L1 = 2H.

96. If the 3-phase balanced source in the given figure delivers 1500 W at a leading power factor of 0.844, then the value of ZL is approximate __________

A. 90∠32.44°
B. 80∠32.44°
C. 80∠ − 32.44°
D. 90∠ − 32.44°

Answer: D

Average power 3VP IP cosθ = 1500

3\((\frac{V_L}{\√{3}}) (\frac{V_L}{\√{3} Z_L})\) cos θ = 1500

Z<sub>L</sub> = \(\frac{V_L^2}{1500} = \frac{400^2 (0.844)}{1500}\) = 90 Ω

θ = ∠ − arc cos⁡(0.844)

= ∠ − 32.44°.

97. An RLC series circuit has a resistance R of 20Ω and a current which lags behind the applied voltage by 45°. If the voltage across the inductor is twice the voltage across the capacitor, the value of inductive resistance is _________

A. 10 Ω
B. 20 Ω
C. 40 Ω
D. 60 Ω

Answer: C

Z = 20 + j20

V = VR = j (VL – VC)

Given, VL = 2 VC

ZL = 2 ZC

ZL – ZC = 20

2 ZC – ZC = 20

ZC = 20 Ω

ZL = 2ZC = 40 Ω.

98. In the Wheatstone bridge shown below, if the resistance in each arm is increased by 0.05%, then the value of Vout will be?

A. 50 mV
B. Zero
C. 5mV
D. 0.1mV

Answer: B

In Wheatstone bridge, balance condition is

R1R3 = R2R4

Here,

R1 = 5, R2 = 10, R3 = 16, R4 = 8

And when the Wheatstone bridge is balanced then, at Vout voltage will be Zero.

99. The voltage drop across a standard resistor of 0.2 Ω is balanced at 83 cm. The magnitude of the current, if the standard cell emf of 1.53 V is balanced at 42 m is ___________

A. 13.04 A
B. 10 A
C. 14.95 A
D. 12.56 A

Answer: C

Voltage drop per unit length = 1.53/42 = 0.036 V/cm

Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V

∴ Current through resistor, I = 2.99/0.2 = 14.95 A.

100. The readings of polar type potentiometer are
I = 12.4∠27.5°
V = 31.5∠38.4°
Then the Reactance of the coil will be?

A. 2.51 Ω
B. 2.56 Ω
C. 2.54 Ω
D. 2.59 Ω

Answer: C

Here, V = 31.5∠38.4°

I = 12.4∠27.5°

Z = (31.5∠38.4°)/(12.4∠27.5°) = 2.54∠10.9°

But

Z = R + jX = 2.49 + j0.48

∴ Reactance X = 2.54 Ω.

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