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

101. 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).

 

102. 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%.

 

103. 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.

 

104. 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.

 

105. 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.

 

106. In the figure given below, a 220 V 50 Hz supplies a 3 − phase balanced source. The pressure Coil (PC. and Current Coil (CC. of a wattmeter are connected to the load as shown. The wattmeter reading is _________

In the figure given below, a 220 V 50 Hz supplies a 3-phase balanced source. The pressure Coil (PC. and Current Coil

A. Zero
B. 1600 W
C. 242 W
D. 400 W

Answer: C

Wattmeter reading = Current through CC × Voltage across PC × cos (phase angle).

IBR = ICC = {220∠120°/100° = 2.2∠120°

VYB  = VPC  = 220∠ − 120°

w = 2.2∠120° × 220∠ − 120° × cos 240° = – 242 W.

 

107. In the Owen’s bridge shown in below figure, Z1  = 200∠60°, Z2  = 400∠ − 90°, Z3  = 300∠0°, Z4  = 400∠30°. Then,

In the Owen’s bridge shown in below figure, Z1 = 200∠60°, Z2 = 400∠-90°, Z3 = 300∠0°, Z4 = 400∠30°. Then,

A. Bridge is balanced with given impedance values
B. Bridge can be balanced, if Z4  = 600∠60°
C. Bridge can be balanced, if Z3  = 400∠0°
D. Bridge cannot be balanced with the given configuration

Answer: D

For Bridge to be balanced, the product of impedances of the opposite arm should be equal in magnitude as well as phase angle.

Here Z3 Z2 ≠ Z1 Z4 for whatever chosen value. Therefore the Bridge cannot be balanced.

 

108. In Maxwell’s capacitance bridge for calculating unknown inductance, the various values at balance are, R1  = 300 Ω, R2  = 700 Ω, R3  = 1500 Ω, C4  = 0.8 μF. The values of R1, L1 and Q factor, if the frequency is 1100 Hz are _________

A. 240 Ω, 0.12 H, 3.14
B. 140 Ω, 0.168 H, 8.29
C. 140 Ω, 0.12 H, 5.92
D. 240 Ω, 0.36 H, 8.29

Answer: B

From Maxwell’s capacitance, we have

R1 = R2 R3/R4 = 300 ×700/1500 = 140 Ω

L1  = R2 R3 C4

= 300 × 700 × 0.8 × 10 − 6  = 0.168 H
∴ Q = ωL1/R1

= 2 × π × 1100 × 0.168/140 = 8.29.

 

109. In the figure below, the values of the resistance R1 and inductance L1 of a coil are to be calculated after the bridge is balanced. The values are _________________

In the figure below, the values of the resistance R1 and inductance L1 of a coil are to be calculated after the bridge is balanced.

A. 375 Ω and 75 mH
B. 75 Ω and 150 mH
C. 37.5 Ω and 75 mH
D. 75 Ω and 75 mH

Answer: A

Applying the usual balance condition relation,

Z1 Z4  = Z2 Z3

We have, (R1 + jL1 ω) \(\frac{R_4/jωC_4}{R_4+1/jωC_4}\) = R2 R3

Or, R1 R4 + jL1 ωR4  = R2 R3 + j R2 R3 R4 C4 ω

∴ R1 = 2000 × 750/4000 = 375 Ω

∴ L1  = 2000 × 750 × 0.5 × 10 − 6  = 75 mH.

 

110. The four arms of an AC bridge network are as follows:

Arm AB: unknown impedance
Arm BC: standard capacitor C2 of 1000pf
Arm CD: a non-inductive resistance of R of 100 Ω in parallel to a capacitor of 0.01 μF
Arm DA: a non-inductive resistance of 1000 Ω

The supply frequency is 50 Hz and connected across terminals B and D. If the bridge is balanced with the above value, determine the value of unknown Impedance.

A. 10 kΩ
B. 100 kΩ
C. 250 kΩ
D. 20 kΩ

Answer: A

For the balance conditions,

Z1 Z3 = Z2 Z4

1000 × \(\frac{1}{jω × 1000 × 10^{ − 12}} = (R + jX) \frac{100}{1 + j100 × ω × 0.01 × 10^{ − 6}}\)

\(\frac{10^{12}}{jω} = (R + jX) \left(\frac{100}{1 + jω + 10^{ − 6}}\right)\)

\(\frac{ − j 10^{10}}{ω}\) – 104 = R + jX

Comparing the real part, we get,

R = 10 kΩ.

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