A galvanometer of resistance `50 Omega ` is connected to a battery of `3 V` along with resistance of `2950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 division the above series resistance should be

Options

A. `5050 Omega`
B. `5550 Omega`
C. `6050 Omega`
D. `4450 Omega`

Answer

D

Explanation

(d) Current through the galvanometer,
`I = (3)/((50 + 2950)) = 10^(-3) A`
Current for 30 divisions `= 10^(-3) A`
Current of 20 division `= (10^(-3))/(30) xx 20 = (2)/(3) xx 10^(-3) A`
For the same deflection to obtain for 20 division, let resistance added be `R`
`:. (2)/(3) xx 10^(-3) = (3)/((50 + 1 R))`
or `R = 4450 Omega`