250+ TOP MCQs on Problems Involving Coupling Coefficient and Answers

Network Theory Multiple Choice Questions on “Problems Involving Coupling Coefficient”.

1. 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
Clarification: 3VP IP cosθ = 1500
Or, 3((frac{V_L}{sqrt{3}}) (frac{V_L}{sqrt{3} Z_L})) cos θ = 1500
Or, ZL = (frac{V_L^2}{1500} = frac{400^2 (0.844)}{1500}) = 90 Ω
And θ = ∠-arc cos⁡(0.844)
= ∠-32.44°.

2. 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
Clarification: Z = 20 + j20
V = VR = j (VL – VC)
Given, VL = 2 VC
Or, ZL = 2 ZC
Or, ZL – ZC = 20
Or, 2 ZC – ZC = 20
Or, ZC = 20 Ω
Or, ZL = 2ZC = 40 Ω.

3. 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
Clarification: 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.

4. 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
Clarification: Voltage drop per unit length = (frac{1.53}{42}) = 0.036 V/cm
Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V
∴ Current through resistor, I = (frac{2.99}{0.2}) = 14.95 A.

5. The readings of polar type potentiometer are
I = 12.4∠27.5°
V = 31.5∠38.4°
Then, Reactance of the coil will be?
A. 2.51 Ω
B. 2.56 Ω
C. 2.54 Ω
D. 2.59 Ω

Answer: C
Clarification: Here, V = 31.5∠38.4°
I = 12.4∠27.5°
Z = (frac{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.

6. The simultaneous applications of signals x (t) and y (t) to the horizontal and vertical plates 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
Clarification: (frac{f_y}{f_x} = frac{x-peak}{y-peak})
Here, x-peak = 1 and y-peak = 2
∴ y(t) = Q sin (2t + 15).

7. 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
Clarification: Here, R1 and R2 are in parallel.
Then, (frac{1}{R} = frac{1}{R_1} + frac{1}{R_2})
Or, R = (frac{50}{15}) kΩ
∴ (frac{△R}{R} = frac{△R_1}{R_1^2} + frac{△R_2}{R_2^2})
And △R1 = 0.5 × 103, △R2 = 0.5 × 103
∴ (frac{△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})
= (frac{0.5}{30} + frac{1}{15} = frac{2.5}{30}) = 8.33%.

8. 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
Clarification: Turn compensation only alters ratio error n=400
Ratio error = -0.5% = – (frac{0.5}{100}) × 400 = -2
So, Actual ratio = R = n+1 = 401
Nominal Ratio KN = (frac{400}{1}) = 400
Now, if the number of turns are reduced by one, n = 399, R = 400
Ratio error = (frac{K_N-R}{R} = frac{200-200}{200}) = 0.

9. 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
Clarification: 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.

10. 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
Clarification: Meter constant = (frac{Number, of, revolution}{Energy} = frac{600 × 230 × 15 × 0.8}{1000}) = 1656
∴ Speed in rpm = (frac{1656}{60}) = 27.6 rpm.

11. 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 watt-meter are connected to the load as shown. The watt-meter reading is _________

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

Answer: C
Clarification: Watt-meter reading = Current through CC × Voltage across PC × cos (phase angle).
IBR = ICC = (frac{220∠120°}{100°}) = 2.2∠120°
VYB = VPC = 220∠-120°
w = 2.2∠120° × 220∠-120° × cos 240° = – 242 W.

12. 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
Clarification: 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.

13. 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
Clarification: From Maxwell’s capacitance, we have
R1 = (frac{R_2 R_3}{R_4} = frac{300 ×700}{1500}) = 140 Ω
L1 = R2 R3 C4
= 300 × 700 × 0.8 × 10-6 = 0.168 H
∴ Q = (frac{ωL_1}{R_1})
= (frac{2 × π × 1100 × 0.168}{140}) = 8.29.

14. 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 _________________

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
Clarification: 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 × (frac{750}{4000}) = 375 Ω
∴ L1 = 2000 × 750 × 0.5 × 10-6 = 75 mH.

15. 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
Clarification: 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}})
Or, (frac{10^{12}}{jω} = (R + jX) left(frac{100}{1 + jω + 10^{-6}}right))
Or, (frac{- j 10^{10}}{ω}) – 104 = R + jX
Comparing the real part, we get,
R = 10 kΩ.

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