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 Z_{L} is approximately __________

A. 90∠32.44°

B. 80∠32.44°

C. 80∠-32.44°

D. 90∠-32.44°

Answer: D

Clarification: 3V_{P} I_{P} cosθ = 1500

Or, 3((frac{V_L}{sqrt{3}}) (frac{V_L}{sqrt{3} Z_L})) cos θ = 1500

Or, Z_{L} = (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 = V_{R} = j (V_{L} – V_{C})

Given, V_{L} = 2 V_{C}

Or, Z_{L} = 2 Z_{C}

Or, Z_{L} – Z_{C} = 20

Or, 2 Z_{C} – Z_{C} = 20

Or, Z_{C} = 20 Ω

Or, Z_{L} = 2Z_{C} = 40 Ω.

3. In the Wheatstone bridge shown below, if the resistance in each arm is increased by 0.05%, then the value of V_{out} will be?

A. 50 mV

B. Zero

C. 5mV

D. 0.1mV

Answer: B

Clarification: In Wheatstone bridge, balance condition is

R_{1}R_{3} = R_{2}R_{4}

Here, R_{1} = 5, R_{2} = 10, R_{3} = 16, R_{4} = 8

And when the Wheatstone bridge is balanced then, at V_{out} 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, R_{1} and R_{2} 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 △R_{1} = 0.5 × 10^{3}, △R_{2} = 0.5 × 10^{3}

∴ (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 K_{N} = (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).

I_{BR} = I_{CC} = (frac{220∠120°}{100°}) = 2.2∠120°

V_{YB} = V_{PC} = 220∠-120°

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

12. In the Owen’s bridge shown in below figure, Z_{1} = 200∠60°, Z_{2} = 400∠-90°, Z_{3} = 300∠0°, Z_{4} = 400∠30°. Then,

A. Bridge is balanced with given impedance values

B. Bridge can be balanced, if Z_{4} = 600∠60°

C. Bridge can be balanced, if Z_{3} = 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 Z_{3} Z_{2} ≠ Z_{1} Z_{4} 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, R_{1} = 300 Ω, R_{2} = 700 Ω, R_{3} = 1500 Ω, C_{4} = 0.8 μF. The values of R_{1}, L_{1} 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

R_{1} = (frac{R_2 R_3}{R_4} = frac{300 ×700}{1500}) = 140 Ω

L_{1} = R_{2} R_{3} C_{4}

= 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 R_{1} and inductance L_{1} 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,

Z_{1} Z_{4} = Z_{2} Z_{3}

We have, (R_{1} + jL_{1} ω) (frac{R_4/jωC_4}{R_4+1/jωC_4}) = R_{2} R_{3}

Or, R_{1} R_{4} + jL_{1} ωR_{4} = R_{2} R_{3} + j R_{2} R_{3} R_{4} C_{4} ω

∴ R_{1} = 2000 × (frac{750}{4000}) = 375 Ω

∴ L_{1} = 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 C_{2} 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,

Z_{1} Z_{3} = Z_{2} Z_{4}

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}) – 10^{4} = R + jX

Comparing the real part, we get,

R = 10 kΩ.