Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

Options

A. `40 Omega`
B. `25 Omega`
C. `250 Omega`
D. `500 Omega`

Answer

C

Explanation

Let N = number of turns is galvanometer
A = Area, B = magnetic field
K = the restoring torque per unit twist.
Current sensitivity , `I_S = (NBA)/(k)`
Voltage sensitivity,
`V_S = (NBA)/(kR_G)`
So, resitance of galvanometer
`R_(G) = (I_s)/(V_s) = (5 xx1)/(20 xx 10^(-3)) = (5000)/(20) = 250 Omega`