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

1. 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 the _______

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

View Answer

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 Z

_{3}Z

_{2}≠ Z

_{1}Z

_{4}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, R_{1} = 300 Ω, R_{2} = 700 Ω, R_{3} = 1500 Ω, C_{4} = 0.8 μF. Calculate R_{1}, L_{1} 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

View Answer

Explanation: 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.

3. In Wein’s bridge, the output frequency is determined by __________

a) RLC combination

b) LC combination

c) RC combination

d) RL combination

View Answer

Explanation: 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. The disadvantage of Maxwell Bridge is?

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

View Answer

Explanation: 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 ∝ L

_{X}

∴ Inductance cannot be measured over a wide range.

5. The Bridge shown in the below figure is ______________

a) Maxwell’s Bridge

b) Wien’s Bridge

c) Anderson’s Bridge

d) Hay’s Bridge

View Answer

Explanation: 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

View Answer

Explanation: 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 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

View Answer

Explanation: 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.

8. Maxwell’s Inductance Capacitance Bridge is used for measuring ___________

a) Inductance

b) Capacitance

c) Frequency

d) Mutual Inductance

View Answer

Explanation: 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 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

View Answer

Explanation: 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, \(\frac{10^{12}}{jω} = (R + jX) \frac{100}{1 + jω + 10^{-6}} \)

Or, \(\frac{- j10^{10}}{ω} – 10^4\) = 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 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 Capacitance.

a) 100 pf

b) 1000 pf

c) 500 pf

d) 10 pf

View Answer

Explanation: 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, \(\frac{10^{12}}{jω} = (R + jX) \frac{100}{1 + jω + 10^{-6}} \)

Or, \(\frac{- j10^{10}}{ω} – 10^4\) = R + jX

Comparing the imaginary part, we get,

\(\frac{1}{ωC} = \frac{10^{10}}{ω}\)

Or, C = 100 pf.

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