JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Current Electricity

A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of

4 \times 1 0^{- 4} A

passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :

(A)250 ohm
(B)200 ohm✓ Correct
(C)6200 ohm
(D)6250 ohm

Answer: (B)

Solution

Full-scale deflection is

25

divisions, so the current there is

i_{g} = 25 \times 4 \times 1 0^{- 4} = 1 0^{- 2}

A. To read up to

2.5

V a series resistance

R

is added so that

V = i_{g} (R_{G} + R)

with

R_{G} = 50 Ω

2.5 = 1 0^{- 2} (50 + R)

, giving

250 = 50 + R

and

R = 200 Ω

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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