Question

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

• A. $500 \, \Omega$
• B. $40 \, \Omega$
• C. $250 \, \Omega$
• D. $25 \, \Omega$

Answer 1

The Correct Option is C

Current sensitivity
$I_S = \frac{NBA}{C}$
Voltage sensitivity
$V_S = \frac{NBA}{CR_G}$
So, resistance of galvanometer
$R_G = \frac{I_S}{V_S} = \frac{5 \times 1}{20 \times 10^{-3}} = \frac{5000}{20} = 250 \, \Omega $

Answer 2

C.S = SI = \(\frac{\theta}{I}\) = (\(\frac{NBA}{K}\))

SI =\(\frac{5div}{mA}\) =(\(\frac{5div}{10-3}\))

V.S = SV = (\(\frac{NBA}{K}\frac{1}{RG}\)) = 20 \(\frac{div}{v}\)

\(\frac{SI}{SV}\) = (\(\frac{NBA}{K}\))/(\(\frac{NBA}{K}\))\(\frac{1}{RG}\) = RG

RG = \(\frac{\frac{5}{10^-3}}{\frac{20}{1}}\)

RG = 250Ω

The correct option is (C).

Answer 3

First, we’ll find out the current sensitivity of the galvanometer.

The current sensitivity of the galvanometer is given by,

I_S = \(\frac{nBA}{c}\)……………………(1)

Where n is the number of turns in the coil of a galvanometer,

B is the magnetic field around the coil.

A is the area of the coil.

c is the restoring torque for 1 unit twist.

It is given that the current sensitivity is 5 \(\frac{div}{mA}\).

If we apply this value in equation (1).

we can write that,

IS=\(\frac{nBA}{c}\) = 5

Equation (2) gives the voltage sensitivity of the galvanometer,

V_S=\(\frac{nBA}{cR}\)

The voltage sensitivity of the galvanometer is \(\frac{20div}{V}\).

Now, we will look for a relation between the voltage sensitivity and the current sensitivity of the galvanometer.

Combining Equations (1) and (2) will give us,

V_S = \(\frac{nBA}{cR}\) = 20

So, the resistance of the galvanometer is,

V_S = \(\frac{I_S}{R}\)

⇒ 20=\(\frac{\frac{5}{10^-3}}{R}\)

⇒ R=250

So, the resistance acquired by the galvanometer is  250 Ω

Hence the correct option is (C).