A galvanometer of resistance ` 50 Omega` gives a full scale deflection for a current `5xx10^(-4) A.`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

Here ,
Resistance of galvanometer , G = 50 `Omega `
Current for full scale deflection , `I_(g) = 5xx 10^(-4) A `
Let R be resistance connected in series with the galvanometer to read 3 V . Then
`V= I_(g) (G+R)`
or `R = V/(I_(g)) - G `
`R = (3V)/(5xx10^(-4)A) - 50 Omega `
` = 6000 Omega - 50 Omega = 5950 Omega `