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:

Step-by-step solution

Given that the galvanometer has 25 divisions and the current required for deflection one division is 4 × 10 - 4   A current for full scale deflection I = 4 × 10 - 4 × 25   A = 10 - 2   A To use this galvanometer as a voltmeter for a maximum voltage of 2 . 5   V , we need to add a resistance R in series to the resistance of the galvanometer of G   =   50   Ω such that, I   =   2 . 5 R + G ⇒ R   =   2 . 5 10 - 2   -   50 = 200