The resistance of a galvanometer is `50 ohm` and the current required to give full scale deflection is `100 muA`. In order to convert it into an ammeter, reading upto `10 A`, it is necessary to put a resistance of

Options

A. `5 xx 10^(-3) Omega` in parallel
B. `5 xx 10^(-4) Omega` in parallel
C. `10^(5) Omega` in series
D. `99,950 Omega` in series

Answer

B

Explanation

(b) Resistance in parallel `S = (Gi_(g))/(i- i_(g)) = (50 xx 100 xx 10^(-6))/((10 - 100 xx 10^(-6)))`
`implies S = 5 xx 10^(-4) Omega`