If the resistance of the galvanometer G is `15 Omega`, then current through the galvanometer will be

Options

A. `1//3A`
B. `1//8A`
C. `1//17A`
D. `1//24A`

Answer

D

Explanation

The equivalent resistance of galvanometer and a resistance of `3 Omega` is ,
`R_(1)=(15xx3)/(15+3)=(45)/(18)=2.5Omega`
Thus, current supplied by cell is,
`I=(2)/(5+2.5+0.5)=2/8=1/4 A`
Now, current through galvanometer is ,
`I_(g)=((3)/(15+3))xx1/4`
`=(1)/(24)A`