A galvanometer coil has a resistance of `100 Omega`. When a current passes through the galvanometer, 1% of the current goes through the coil and the rest through the shunt. Find the resistance of the shunt.

Explanation

The situation is shown in figure. As the potential differences the `100 Omega` coil and across the shunt `R_s` are the same,
`0.01 I xx 100 Omega = 0.99 I xx R_s`
or, R_s = (0.01 xx 100)/(0.99) Omega = 100/99 Omega`.