GATE 2025 IN Q.56: solution and answer

GATE 2025 · INQ.56 · 2 marks · NAT

Consider an AC bridge shown in the figure with $R = 300 Ω$ , $R 1 = 1000 Ω$ , $R 2 = 500 Ω$ , $L = 30 mH$ , and a detector D. At the bridge balance condition, the frequency of the excitation source $V_{S}$ is ________ kHz (rounded off to two decimal places).

GATE 2025 IN question 56: A four-arm bridge drawn as a diamond with top, left, right and bottom vertices. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

At balance $ω L = R$ , so $ω = 1 0^{4} rad/s$ and $f = 1 0^{4} / (2 π) \approx 1.59 kHz$ .

Official answer: 1.55 to 1.65Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Name the arms (top, left, right and bottom vertices of the diamond). Top-left arm, series R and L: $Z_{1} = R + j ω L$ . Top-right arm, R in parallel with L: $Z_{2} = \frac{j ω L R}{R + j ω L}$ . Bottom-left arm: $R_{1}$ . Bottom-right arm: $R_{2}$ . The source is across the top and bottom vertices, and D is across the left and right vertices.
Balance condition: D carries no current when the left and right vertices are at the same potential, i.e. $\frac{R _{1}}{Z _{1} + R _{1}} = \frac{R _{2}}{Z _{2} + R _{2}}$ , which simplifies to $Z_{1} R_{2} = Z_{2} R_{1}$ (products of opposite arms equal). Hence $\frac{Z _{1}}{Z _{2}} = \frac{R _{1}}{R _{2}} = \frac{1000}{500} = 2$ .
Form the ratio: $\frac{Z _{1}}{Z _{2}} = \frac{( R + j ω L ) ^{2}}{j ω L R} = \frac{R ^{2} - ω ^{2} L ^{2} + j 2 ω L R}{j ω L R} = 2 - j \frac{R ^{2} - ω ^{2} L ^{2}}{ω L R}$
Equate this to $2 + j 0$ . The real part is already $2$ , so the resistance-ratio condition holds at every frequency. The imaginary part must vanish: $R^{2} = ω^{2} L^{2}$ , so (taking the positive root) $ω L = R$ .
$ω = \frac{R}{L} = \frac{300}{30 \times 1 0 ^{- 3}} = 1 0^{4} rad/s$
$f = \frac{ω}{2 π} = \frac{1 0 ^{4}}{2 π} = 1591.55 Hz \approx 1.59 kHz$

KEY IDEA

With a series RL arm $Z_{1}$ and a parallel RL arm $Z_{2}$ made of the same R and L, $Z_{1} / Z_{2} = (R + j ω L)^{2} / (j ω L R)$ always has real part 2. The resistor ratio $R_{1} / R_{2} = 2$ therefore holds at every frequency, and balance only needs the imaginary part to vanish, which gives $ω L = R$ . The bridge balances at only one frequency, so it can be used to measure frequency.

WATCH OUT FOR

Reading the top-right arm as a second series RL branch: then $Z_{1} = Z_{2}$ , the ratio is 1 rather than 2, and the bridge can never balance. Other slips: pairing adjacent arms ( $Z_{1} R_{1} = Z_{2} R_{2}$ ), which gives a ratio of 1/2 that the real part 2 can never match; stopping at $ω = 1 0^{4} rad/s$ without dividing by $2 π$ ; and entering 1591.55 (in Hz) when the question asks for kHz.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 14 Bridges questions, with notes

GATE 2025 IN Q.56: solution and answer

Answer and solution

More Bridges questions