In an ammeter 10% of main current is passing through the galvanometer. If the resistance of the galvanometer is G, then the shunt resistance, in ohms is

Options

A. `60 omega`
B. `240 omega`
C. `120 omega`
D. `480 omega`

Answer

C

Explanation

Shunt is a low resistance connected in parallel with the galvanometer or ammeter

shnt is given by `S=(l_(g)G)/(l-l_(g))`
`therefore 5=((4/100)iG)/(i-4/100l)=(4G)/(96)`
`rarr G=(96xx5)/(4)=120 omega`