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 A. `5 xx 10^(-3) Omega` in parallelB. `5 xx 10^(-4) Omega` in parallelC. `10^(5) Omega` in seriesD. `99,950 Omega` in series

The resistance of a galvanometer is `50 ohm` and the current required

1.	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 A. `5 xx 10^(-3) Omega` in parallelB. `5 xx 10^(-4) Omega` in parallelC. `10^(5) Omega` in seriesD. `99,950 Omega` in series
Answer» Correct Answer - B (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`

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