Question

In an ammeter, 4% of the main current is passing through the galvanometer. If shunt resistance is 5 `Omega`, then resistance of galvanometer will be
A. `60Omega`
B. `120Omega`
C. `240Omega`
D. `480Omega`

Answer 1

Correct Answer - B
`(I_g)/(I)=4%,S=5Omega,G=?`
`(I_g)/(I)=(S)/(S+G)`
`(4)/(100)=(5)/(5+G)`
`(1)/(25)=(5)/(5+G)`
`5+G=125`
`G=120Omega `.