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

Resistance of galvanometer,
`R_(1) = 100 Omega`
Connected resistance,
`R_(2) = 10 k Omega = 10000 Omega`
Total resistance,
`R = 10000 + 100`
`= 10100 Omega`
Current of galvanometer,
`i_(g) = (V)/(R)`
`= (10)/(10100)`
`= 9.9 xx 10^(-4)`
Similary, `S = (i_(g))/(i-i_(g)) xx 100`
`= (i_(g))/((1-10 xx 10^(-4))) xx 100`
`= (100)/(0.99) xx 10 xx 10^(-4) = 0.1 Omega`