Mutual Inductance MCQ [Free PDF] – Objective Question Answer for Mutual Inductance Quiz

21. 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 __________

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 approximately

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°.

 

22. 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.

 

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

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.

 

24. 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.

 

25. 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.

 

26. The simultaneous applications of signals x (t) and y (t) to the horizontal and vertical planes respectively, of an oscilloscope, produce a vertical figure of 8 displays. If P and Q are constants and x(t) = P sin (4t + 30), then y(t) is equal to _________

A. Q sin (4t – 30)
B. Q sin (2t + 15)
C. Q sin (8t + 60)
D. Q sin (4t + 30)

Answer: B

\(\frac{f_y}{f_x} = \frac{x − peak}{y − peak}\)

Here, x − peak = 1 and y − peak = 2

∴ y(t) = Q sin (2t + 15).

 

27. A resistor of 10 kΩ with a tolerance of 5% is connected in parallel with 5 kΩ resistors of 10% tolerance. The tolerance limit is _______

A. 9%
B. 12.4%
C. 8.33%
D. 7.87%

Answer: C

Here, R1 and R2 are in parallel.

Then

\(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)

R = 50/15 kΩ

∴ △R/R = \frac{△R_1}{R_1^2} + \frac{△R_2}{R_2^2}\)

And △R1  = 0.5 × 103, △R2  = 0.5 × 103

∴ △R/R = \frac{10 × 10^3}{3 × 10 × 10^3} × \frac{0.5 × 10^3}{10 × 10^3} + \frac{10}{3} × \frac{10^3}{5 × 10^3} × \frac{0.5 × 10^3}{5 × 10^3}\)

= 0.5/30 + 1/15 = 2.5/30  = 8.33%.

 

28. A 200/1 Current Transformer (CT) is wound with 400 turns on the secondary on a toroidal core. When it carries a current of 180 A on the primary, the ratio is found to be − 0.5%. If the number of secondary turns is reduced by 1, the new ratio error (in %) will be?

A. 0.0
B. − 0.5
C. − 1.0
D. − 2.0

Answer: C

Turn compensation only alters ratio error n = 400

Ratio error = − 0.5% = – 0.5/100 × 400 = − 2

So, Actual ratio = R = n+1 = 401

Nominal Ratio KN = 400/1 = 400

Now, if the number of turns are reduced by one, n = 399, R = 400

Ratio error = (KN − R)/R

= 200 − 200/200 = 0.

 

29. In the Two wattmeter method of measuring power in a balanced three-phase circuit, one wattmeter shows zero and the other positive maximum. The load power factor is ___________

A. 0
B. 0.5
C. 0.866
D. 1.0

Answer: B

The load power factor is = 0.5. Since at this power factor one wattmeter shows zero and the other shows a positive maximum value of power.

 

30. 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

Answer: C

Meter constant = Number of revolution/Energy

= 600 × 230 × 15 × 0.8/1000 = 1656

∴ Speed in rpm = 1656/60 = 27.6 rpm.

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