A galvanometer of resistance 50 `Omega` gives a full scale deflection for a current `5 xx 10^(-4)` A. The resistance that should be connected in series with the galvanometer to read 3 V is

Options

A. `5050 Omega`
B. `5950 Omega`
C. `595 Omega`
D. `5059 Omega`

Answer

B

Explanation

`R_(G)= 50 Omega, I_(g) = 5 xx 10^(-4) A`
`V= I_(g) (R + R_(G))`
`rArr R = V/I_(g) - R_(G) = 3/(5 xx 10^(-4)) - 50`
`= 6000 - 50 = 5950 Omega`