To send `10%` of the main current through a moving coil galvanometer of resistance `9Omega`, shunt S required is

Options

A. `81Omega`
B. `1Omega`
C. `10Omega`
D. `9Omega`

Answer

B

Explanation

For Parallel connection, potential difference across the node is same, `i_g xx G = (i- i_g) xx S,` where `i_g`, G, i&S represent current in galvanometer, galvanometer resistance, total current and shunt resistance respectively.
`rArr S = (0.1)/(0.9) xx G { i_g = 0.1i}`
`rArr S = 1/9 xx 9 = 1 Omega`