Question

A galvanometer with 50 divisions on the scale has a resistance of 25 Ω. A current of 2 x 10^-4 A gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to 25 V is

(a) 1200 Ω

(b) 1000 Ω

(c) 2475 Ω

(d)  2500 Ω

Answer 1

Correct option  (C)  2475Ω

Explanation:

I = V/R + G

Total current through galvanometer

= 50 × 2 ×10^–4 = 10^–2A 

We know that, I = V/R + G

⇒ 10^–2(25 + R) = 25 

⇒ 25 + R = 2500

 R = 2475 Ω