GATE 2024 IN Q.54: solution and answer

GATE 2024 · INQ.54 · 2 marks · NAT

Consider the figure shown. For zero deflection in the galvanometer, the required value of resistor $R_{x}$ is ____ $Ω$ (rounded off to nearest integer).

GATE 2024 IN question 54: A diamond-shaped bridge with five nodes: top node T, left node L, centre node C, right node R and bottom node B. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

$R_{x} = 60 Ω$

Official answer: 58 to 62Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Label the nodes: top $T$ , left $L$ , centre $C$ , right $R$ , bottom $B$ . The source $V_{in}$ is connected between $L$ (+) and $R$ (−). Take $V_{R} = 0$ , so $V_{L} = V_{in}$ .
Zero galvanometer deflection means no current in the galvanometer branch $T$ – $C$ , so $V_{T} = V_{C}$ . Because that branch carries no current, treat it as open when you find the node voltages.
Look at the lower network. $C$ and $B$ each connect to $L$ through $40 Ω$ and to $R$ through $60 Ω$ . That is a balanced bridge of its own, so $V_{C} = V_{B}$ and the $60 Ω$ resistor between $C$ and $B$ carries no current. (Check: with that resistor removed, both nodes sit at $0.6 V_{in}$ , so it really does carry no current, and the solution is unique.)
Voltage divider along $L$ – $C$ – $R$ : $V_{C} = V_{in} \frac{60}{40 + 60} = 0.6 V_{in}$
Voltage divider along $L$ – $T$ – $R$ (no current leaves $T$ through the galvanometer): $V_{T} = V_{in} \frac{90}{R _{x} + 90}$
Balance condition $V_{T} = V_{C}$ : $\frac{90}{R _{x} + 90} = 0.6 \Rightarrow R_{x} + 90 = 150 \Rightarrow R_{x} = 60 Ω$ This is the usual Wheatstone ratio $\frac{R _{x}}{90} = \frac{40}{60}$ .

KEY IDEA

With zero galvanometer current, nodes $T$ and $C$ are at the same potential. The lower network ( $40 Ω$ and $60 Ω$ from $L$ to $R$ through both $C$ and $B$ ) is itself balanced, so the $60 Ω$ resistor from $C$ to $B$ carries no current. $V_{C}$ is then just the $40 : 60$ divider, and the balance reduces to the ratio $R_{x} /90 = 40/60$ .

WATCH OUT FOR

Setting up the ratio with the arms swapped, for example $R_{x} /90 = 60/40$ , which gives $135 Ω$ . Another mistake is assuming the bottom node $B$ is wired to the source, or trying to solve the full five-node network when the $C$ – $B$ resistor carries no current. The arms that must be in the same ratio are the ones that meet at the source terminals: $R_{x}$ and $40 Ω$ both touch $L$ , and $90 Ω$ and $60 Ω$ both touch $R$ .

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2024 IN Q.54: solution and answer

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