250+ TOP MCQs on Advanced Problems on Measurement of Inductance using AC Bridges and Answers

Tough Electrical Measurements Questions and Answers on “Advanced Problems on Measurement of Inductance using AC Bridges”.

1. In the Owen’s bridge shown in below figure, Z1 = 200∠60°, Z2 = 400∠-90°, Z3 = 300∠0°, Z4 = 400∠30°. Then the _______
tough-electrical-measurements-questions-answers-q1
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.

2. 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. Calculate R1, L1 and Q factor, if the frequency is 1100 Hz.
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.

3. In Wein’s bridge, the output frequency is determined by __________
a) RLC combination
b) LC combination
c) RC combination
d) RL combination
Answer: c
Clarification: The frequency of Wien Bridge is given by
f = (frac{1}{2π(R_1 R_2 C_1 C_2 )^{0.5}}) Hz
∴ The output frequency is determined by the RC combinations.

4. What is the disadvantage of Maxwell Bridge?
a) Inductance cannot be measured over a wide range
b) Measurement is not independent of frequency
c) Number of components is large
d) Inductance can be measured over a wide range
Answer: a
Clarification: Maxwell Bridge cannot be used for the measurement of high Q values.
We have, Q = (frac{1}{ωR_X C_X} = frac{ωL_X}{R_X})
Hence, Q ∝ LX
∴ Inductance cannot be measured over a wide range.

5. The Bridge shown in the below figure is ______________
tough-electrical-measurements-questions-answers-q5
a) Maxwell’s Bridge
b) Wien’s Bridge
c) Anderson’s Bridge
d) Hay’s Bridge
Answer: a
Clarification: The given figure is Maxwell’s Bridge because it consists of two inductors. It is also used to measure the inductance of the inductor.

6. Hay’s Bridge is used for measuring __________
a) Resistance in the milliohm range
b) Low values of capacitance
c) Comparison of resistances which are nearly equal
d) The Inductance of a coil with a large time constant
Answer: d
Clarification: Carey – Foster slide-wire is suited for the comparison of resistances which are nearly equal. The Schering Bridge is suited for Low values of capacitance. The Kelvin double bridge is used for measuring resistance in the milliohm range. Hay’s Bridge is suited for the measurement of Inductance of a coil with a large time constant.

7. In Maxwell’s Bridge, as shown 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?
tough-electrical-measurements-questions-answers-q7
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.

8. Maxwell’s Inductance Capacitance Bridge is used for measuring ___________
a) Inductance
b) Capacitance
c) Frequency
d) Mutual Inductance
Answer: a
Clarification: For measuring Capacitance De-Sauty’s Bridge and Schering Bridge should be used. For measuring Frequency Wien’s Bridge is used. For measuring Mutual Inductance Heaviside and Campbell’s Bridge are used.

9. 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) (frac{100}{1 + jω + 10^{-6}} )
Or, (frac{- j10^{10}}{ω}) – 104 = R + jX
Comparing the real part, we get,
R = 10 kΩ.

10. 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 Capacitance.
a) 100 pf
b) 1000 pf
c) 500 pf
d) 10 pf
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) (frac{100}{1 + jω + 10^{-6}} )
Or, (frac{- j10^{10}}{ω}) – 104 = R + jX
Comparing the imaginary part, we get,
(frac{1}{ωC} = frac{10^{10}}{ω})
Or, C = 100 pf.

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