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

Current through the galvanometer
`l=(3)/((50+2950))=10^(-3)A`

Currect for 30 divisions `=10^(-3)A`
Current for 20 divisions `=(10^(-3))/(30)xx20`
`=(2)/(3)xx10^(-3)A`
For the same deflection to obtain for 20 divisions, let resistance added be R
`:." "(2)/(3)xx10^(-3)=(3)/((50+1R))`
or `" "R=4450 Omega`