A moving coil galvanometer is converted into an ammeter reads up to `0.03 A` by connecting a shunt of resistance `4r` across it and ammeter reads up `0.06 A`, when a shunt of resistance `r` is used. What is the maximum current which can be sent through this galvanometer if no shunt is used ?

Options

A. `0.03 A`
B. `0.04 A`
C. `0.02 A`
D. `0.01 A`

Answer

C

Explanation

(a) Let `G` be the resistance of galvanometer and `i_(g)` the current which on passing through the galvanomete produces full-scale deflection.
Since `G` and `S` in parallel, the potential difference across them will be
`i_(g) xx G = (I - i_(1)) xx S`
`(i_(g))/(i) = (S)/(S + G)`
Hence, `i_(g) = (4 r)/(4r + G) xx 0.03`
`i_(g) = (r )/(r + G) xx 0.06`
Equaiting Eqs. (1) and (2), we get
`4r + G = 2 (r + G)`
`implies G = 2r`
`:. i_(g) = (4r)/(4r + 2r) xx 0.03 = 0.02 A` [from Eq. (1)]