GATE 2022 IN Q.63: solution and answer

GATE 2022 · INQ.63 · 2 marks · NAT

The bridge shown is balanced when $R_{1} = 100 Ω$ , $R_{2} = 210 Ω$ , $C_{2} = 2.9 μ F$ , and $R_{4} = 50 Ω$ . The 2 kHz sine-wave generator supplies a voltage of $10 V_{p-p}$ . The value of $L_{3}$ (in millihenry) is ______ (round off to two decimal places)

GATE 2022 IN question 63: Four-arm AC bridge (Maxwell inductance-capacitance bridge). — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

$L_{3} = R_{1} R_{4} C_{2} = 100 \times 50 \times 2.9 μ F = 14.50 mH$ .

Official answer: 14.40 to 14.60Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Identify the arms. The detector sits between B and D and the generator between A and C, so $Z_{A B} = R_{1}$ , $Z_{B C} = Z_{2} = R_{2} ∥ C_{2}$ , $Z_{A D} = Z_{3} = R_{3} + j ω L_{3}$ and $Z_{D C} = R_{4}$ .
Balance condition. At balance $V_{B} = V_{D}$ , so $\frac{Z _{A B}}{Z _{A B} + Z _{B C}} = \frac{Z _{A D}}{Z _{A D} + Z _{D C}}$ . This reduces to the product of opposite arms, $Z_{A B} Z_{D C} = Z_{B C} Z_{A D}$ , that is $R_{1} R_{4} = Z_{2} Z_{3}$ .
Write the parallel arm as an admittance: $\frac{1}{Z _{2}} = \frac{1}{R _{2}} + j ω C_{2}$ . Then $Z_{3} = \frac{R _{1} R _{4}}{Z _{2}} = R_{1} R_{4} (\frac{1}{R _{2}} + j ω C_{2})$
Equate real and imaginary parts of $R_{3} + j ω L_{3}$ : $R_{3} = \frac{R _{1} R _{4}}{R _{2}}$ and $ω L_{3} = ω R_{1} R_{4} C_{2}$ , so $L_{3} = R_{1} R_{4} C_{2}$ . The $ω$ cancels, which is the defining property of the Maxwell inductance-capacitance bridge.
Compute: $L_{3} = (100) (50) (2.9 \times 1 0^{- 6}) = 1.45 \times 1 0^{- 2} H = 14.50 mH$
Check: $R_{3} = \frac{100 \times 50}{210} = 23.81 Ω$ , which is positive and physically sensible, but it is not asked for. The 2 kHz frequency and the $10 V_{p-p}$ amplitude do not enter either balance equation. The frequency cancels, and the amplitude only affects how sensitive the detector is near balance.

KEY IDEA

This is a Maxwell inductance-capacitance bridge. The unknown $R_{3} + j ω L_{3}$ arm is opposite the parallel $R_{2} ∥ C_{2}$ arm. Balance gives $R_{1} R_{4} = Z_{2} Z_{3}$ , so $Z_{3} = R_{1} R_{4} (\frac{1}{R _{2}} + j ω C_{2})$ . That yields $L_{3} = R_{1} R_{4} C_{2}$ and $R_{3} = R_{1} R_{4} / R_{2}$ . Both are independent of frequency, so the 2 kHz and $10 V_{p-p}$ data are distractors.

WATCH OUT FOR

Pairing the wrong arms. Using adjacent arms, for example $L_{3} = R_{1} R_{2} C_{2}$ , gives $60.9 mH$ . The capacitance multiplies the two resistors that form the other pair of opposite arms, here $R_{1}$ and $R_{4}$ . A second trap is treating $C_{2}$ as in series with $R_{2}$ (Hay's bridge). That brings in $ω = 2 π \times 2000 rad/s$ and gives about $0.24 mH$ . A third is quoting $0.0145$ , the value in henry, when the answer is asked for in millihenry.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 14 Bridges questions, with notes

GATE 2022 IN Q.63: solution and answer

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