A galvanometer of resistance `100 Omega` is converted to a voltmeter of range 10 V by connecting a resistance of `10 k Omega`. The resistance required to convert the same galvanometer to an ammeter of range 1 A is

Options

A. `0.4 Omega`
B. `0.3 Omega`
C. `1.2 Omega`
D. `0.1 Omega`

Answer

D

Explanation

`I_(g) =10/(100 + 10 xx 10^(3)) = 1 mA`
`I = 1A, S = (I_(g)/(I-I_(g)))G`
`= (10^(-3) xx 100)/(1- 10^(-3)) rArr S = 0.1 Omega`