GATE 2018 IN Q.49: solution and answer

GATE 2018 · INQ.49 · 2 marks · NAT

The inductance of a coil is measured using the bridge shown in the figure. Balance ( $D = 0$ ) is obtained with $C_{1} = 1 nF$ , $R_{1} = 2.2 M Ω$ , $R_{2} = 22.2 k Ω$ , $R_{4} = 10 k Ω$ . The value of the inductance $L_{x}$ (in mH) is ______.

GATE 2018 IN question 49: A four-arm bridge drawn as a diamond with top, left, right and bottom nodes. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

$L_{x} = R_{2} R_{4} C_{1} = 222 mH$ .

Official answer: 221 to 223Source: Official answer key, GATE 2018 IN (IIT Guwahati)

Name the arms. The source $V_{s}$ is connected between the top and bottom nodes and the detector $D$ between the left and right nodes. $Z_{1} = R_{1} ∥ C_{1}$ (top to left), $Z_{2} = R_{2}$ (top to right), $Z_{4} = R_{4}$ (left to bottom), $Z_{x} = R_{x} + j ω L_{x}$ (right to bottom). So $Z_{1}$ is opposite $Z_{x}$ , and $Z_{2}$ is opposite $Z_{4}$ .
Balance condition. $D = 0$ means the left and right nodes are at the same potential: $\frac{Z _{4}}{Z _{1} + Z _{4}} = \frac{Z _{x}}{Z _{2} + Z _{x}}$ . This simplifies to $Z_{1} Z_{x} = Z_{2} Z_{4}$ (the products of opposite arms are equal). Therefore $Z_{x} = \frac{R _{2} R _{4}}{Z _{1}} = R_{2} R_{4} Y_{1}$ .
Admittance of the parallel $R_{1} C_{1}$ arm: $Y_{1} = \frac{1}{R _{1}} + j ω C_{1}$ .
Substitute, then equate the real parts and the imaginary parts: $R_{x} + j ω L_{x} = \frac{R _{2} R _{4}}{R _{1}} + j ω R_{2} R_{4} C_{1} \Rightarrow L_{x} = R_{2} R_{4} C_{1}, R_{x} = \frac{R _{2} R _{4}}{R _{1}}$ Neither result depends on the frequency. This is the Maxwell inductance-capacitance bridge.
Inductance: $L_{x} = (22.2 \times 1 0^{3} Ω) (10 \times 1 0^{3} Ω) (1 \times 1 0^{- 9} F) = 0.222 H = 222 mH$
Check: $R_{x} = \frac{( 22.2 \times 1 0 ^{3} ) ( 10 \times 1 0 ^{3} )}{2.2 \times 1 0 ^{6}} Ω \approx 100.9 Ω$ , which is a reasonable winding resistance. $R_{1}$ affects only $R_{x}$ and does not appear in $L_{x}$ .

KEY IDEA

This is a Maxwell inductance-capacitance bridge. At balance, $Z_{x} = R_{2} R_{4} Y_{1}$ . The parallel $R_{1} C_{1}$ arm contributes an admittance, so its $j ω C_{1}$ term turns directly into $j ω L_{x}$ . That gives $L_{x} = R_{2} R_{4} C_{1}$ , independent of frequency and of $R_{1}$ .

WATCH OUT FOR

One mistake is using the impedance of the $R_{1} C_{1}$ arm instead of its admittance. Another is pairing the wrong arms, for example writing $Z_{1} Z_{4} = Z_{2} Z_{x}$ when the detector's position makes $Z_{1}$ opposite $Z_{x}$ . A third is bringing $R_{1}$ into the inductance formula. Each of these gives a wrong or frequency-dependent $L_{x}$ . The unit prefixes also cause slips: $22.2 k Ω \times 10 k Ω \times 1 nF = 0.222 H$ , which is 222 mH, not 222 H or 22.2 mH.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 14 Bridges questions, with notes

GATE 2018 IN Q.49: solution and answer

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