A moving coil galvanometer of resistance `20 Omega` gives a full scale deflection when a current of `1mA` is passed through it. It is to be converted into an ammeter reading `20 A` on full scale. But the shunt of `0.005 Omega` only is available. What resistance should be connected in series with the galvanometer coil?

Answer

D

Explanation

In parallel current distributes in inverse ratio of resistance

`:. ((I - i_g)/i_g)=(G+R)/S`
`:. R=((i-i_g)/i_g)S-G`
`=(20/10^-3)(0.005)-20=80Omega`