A galvanometer of resistance `50 Omega` is connected to a battery of 8 V along with a resistance of `3950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 15 divisions, the resistance in series should be .... `Omega`.

Options

A. 7900
B. 1950
C. 2000
D. 7950

Answer

B

Explanation

30 division = 8V
15 division = 4V
`8V = I_(g)(R_(1) + G) =I_(g)(3950 + 50)`
`4V = I_(g) (R_(2) +G) = I_(g) (R_(2) + 50)`
`2 = 4000/(R_(2) + 50) rArr 2R_(2) + 100 = 4000`
`rArr R_(2) = 3900/2= 1950 Omega`