A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × 10-4 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:

Question

A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 × 10-4 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) 6250 ohm (B) 250 ohm (C) 200 ohm (D) 6200 ohm

Tardigrade Admin · Accepted Answer

Ig = 4 × 10-4 × 25 = 10-2 A 2.5 = (50+R) 10-2 ∴ R = 200 Ω

Q. A galvanometer, whose resistance is $50 o hm$ , 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:

6250 ohm

250 ohm

200 ohm

6200 ohm

$I_{g} = 4 \times 1 0^{- 4} \times 25 = 1 0^{- 2} A$
$2.5 = (50 + R) 1 0^{- 2} ∴ R = 200Ω$

Q. A galvanometer, whose resistance is 50ohm, has 25 divisions in it. When a current of 4×10−4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5V, it should be connected to a resistance of: