JEE Main 2025 Jan 23 Shift 2, Physics Q12: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

JEE Main2025Jan 23, Shift 2Physics

Q.

A galvanometer having a coil of resistance $30 Ω$ need $20$ mA of current for full-scale deflection. If a maximum current of $3$ A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} Ω$ , where $X$ is

A
$447$
B
$298$
C
$149$
D
$596$

Solution

In a shunted galvanometer, the same voltage appears across $G$ and $r_{s}$ :

$I_{g} \cdot G = (I - I_{g}) \cdot r_{s} .$

Insert numbers ( $I_{g} = 20 m A = 0.02 A$ , $G = 30 Ω$ , $I = 3 A$ ):

$0.02 \times 30 = (3 - 0.02) \cdot r_{s} \Rightarrow 0.6 = 2.98 r_{s} \Rightarrow r_{s} = \frac{0.6}{2.98} Ω.$

Setting this equal to $\frac{30}{X}$ : $X = \frac{30 \times 2.98}{0.6} = 149.$

Concept behind this question

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