JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Current Electricity

A galvanometer having a coil of resistance

30Ω

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

(A)447
(B)298
(C)149✓ Correct
(D)596

Answer: (C)

Solution

For a shunted galvanometer, the potential difference across the galvanometer coil equals that across the shunt:

I_{g} R_{g} = (I - I_{g}) r_{s}

With

I_{g} = 20 mA = 0.02 A

R_{g} = 30 Ω

and

I = 3 A

(0.02) (30) = (3 - 0.02) r_{s}

0.6 = 2.98 r_{s}

r_{s} = \frac{0.6}{2.98} = \frac{30}{149} Ω

Comparing with

\frac{30}{X}

gives

X = 149

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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