A moving coil galvnometer is converted into an ammeter reading up to `0.03 A` by connecting a shunt of resistance `r/4`. What is the maximum current which can be sent through this galvanometer if no shunt is used (here r=resistance of galvanometer)

Options

A. `0.004A`
B. `0.005A`
C. `0.006A`
D. `0.008A`

Answer

C

Explanation

In parallel current distributes in inverse ratio of resistance.
`(0.03-I_G)/I_G=G/S=r/((r//4))=4`
Solving this equation , we get
`I_G=0.006A`