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

Answer

B

Explanation

Total initial resistance
`=G + R= 50 Omega + 2950 Omega = 3000 Omega`
Current, `I=(3V)/(3000 Omega) = 1 xx 10^(-3) A = 1 mA`
If the deflection has to reduced to 20 divisons, then current
`I' = (1 mA)/(30) xx 20 = 2/3 ma`
Let x be the effective resistance of the circuit,
`3V = 3000 Omega xx 1 mA = x Omega xx 2/3 mA`
or `x = 3000 xx 1 xx 3/2 = 4500 Omega`
`therefore` Resistance to be added `=(4500 Omega - 50 Omega) = 4450 Omega`