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

CONCEPT:

Current sensitivity- Current sensitivity is defined as the deflection in the coil of the galvanometer per unit of the flow of electric current in the circuit and it is defined as;

\({I_s} = \frac{{NBA}}{k}\) -----(1)

Here, N is the number of turns present in the coil, B is the magnetic field, A is the area and k is the restoring torque.

Voltage sensitivity- Voltage sensitivity is defined as the deflection in the galvanometer when a unit voltage is applied in the two terminals of the given circuit and it is defined as;

\({V_s} = \frac{{NBA}}{{k{R_G}}}\) -----(2)

CALCULATION:

Given:

I_s = 5 div/A

 V_s = 20 div/V

Now, on dividing the equation (2) by (1) we get;

\(\frac {I_s}{V_s} = R_G\)

So, the resistance of galvanometer we have,

\({R_G} = \frac{{{I_s}}}{{{V_s}}} = \frac{{5 \times 1}}{{20 \times {{10}^{ - 3}}}} \)

⇒ \(R_G = \frac{{5000}}{{20}}\)

R_G = 250 Ω 

Hence, option 4) is the correct answer.