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. `1000 Omega`
B. `1200 Omega`
C. `1500 Omega`
D. `1800 Omega`

Answer

C

Explanation

Total resistance`=2950+50=3000Omega therefore I_g=3/(3000)=10^(-3)A` To reduce the deflection from 30 div. to 20 div., the current should be 2/3 of the original current. Hence,the resistance should be `3/2 xx 3000=4500 Omega`. `therefore`Additional resistance`=1500Omega`.