GATE 2023 IN Q.30: solution and answer

GATE 2023 · INQ.30 · 1 mark · NAT

When the bridge given below is balanced, the current through the resistor $R_{a}$ is ______ mA (rounded off to two decimal places).

GATE 2023 IN question 30: Kelvin-double-bridge-type network drawn as a rounded loop. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The current through $R_{a}$ is $1.00 mA$ .

Official answer: 0.99 to 1.01Source: Official answer key, GATE 2023 IN (IIT Kanpur)

At balance the galvanometer carries no current: $I_{g} = 0$ . The inner node $G$ connects only to the galvanometer, $R_{a}$ and $R_{b}$ , so by KCL at $G$ the current in $R_{a}$ equals the current in $R_{b}$ .
Because $I_{g} = 0$ , $R_{a}$ and $R_{b}$ form a single series path between the bottom junctions $P$ and $Q$ : $R_{a} + R_{b} = 10 + 10 = 20 m Ω$ . This path is in parallel with the link $R_{c} = 10 m Ω$ , which also joins $P$ and $Q$ .
The $3 mA$ in the lower-right arm arrives at $Q$ . Nothing else is attached to $Q$ except $R_{c}$ and $R_{b}$ , so the whole $3 mA$ divides between these two parallel paths and recombines at $P$ , then continues through the lower-left arm towards ground.
Current divider: each path takes a share proportional to the resistance of the other path. $I_{R_{a}} = I \frac{R _{c}}{R _{c} + ( R _{a} + R _{b} )} = 3 mA \times \frac{10}{10 + 20}$
$I_{R_{a}} = 3 \times \frac{1}{3} = 1.00 mA$ The other $2 mA$ flows through $R_{c}$ (check: $2 + 1 = 3 mA$ ). Once the bridge is balanced, the $1 V$ source, the unlabelled outer arms and the $100 k Ω$ galvanometer resistance do not affect the answer.

KEY IDEA

In a balanced Kelvin double bridge the galvanometer current is zero. The inner arms $R_{a}$ and $R_{b}$ therefore carry the same current and act as one series path of $R_{a} + R_{b}$ , in parallel with the link $R_{c}$ . The $3 mA$ in the arm feeding the link splits by a simple current divider: $I_{R_{a}} = 3 \times \frac{10}{10 + 20} = 1 mA$ .

WATCH OUT FOR

One mistake is to put $R_{a}$ in parallel with $R_{c}$ alone and forget that $R_{b}$ is in series with it, which gives $1.5 mA$ . Another is to flip the divider and use the path's own resistance in the numerator, $3 \times \frac{20}{30} = 2 mA$ , which is actually the current through $R_{c}$ . A third is trying to solve the whole network with the $1 V$ source and the $100 k Ω$ galvanometer. At balance neither affects the answer, and the outer arms have no values printed anyway.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 14 Bridges questions, with notes

GATE 2023 IN Q.30: solution and answer

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