# Network Theory Questions and Answers – Problems Involving Coupling Coefficient

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses 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°

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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.
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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)

Explanation: $$\frac{f_y}{f_x} = \frac{x-peak}{y-peak}$$
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%

Explanation: 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

Explanation: 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

Explanation: 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

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.

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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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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