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). 250 Ω
(B). 40 Ω
(C). 500 Ω
(D). 25 Ω

Answer 1

Hint: Check the formula of current sensitivity and voltage sensitivity of the galvanometer. Current sensitivity and voltage sensitivity depend on the resistor of the galvanometer. We can find the ratio of the sensitivities to find the resistor.

Formula Used:
Current sensitivity of a galvanometer is given by,
$I_S={\displaystyle \frac{nBA}{c}}$……………………(1)

Voltage sensitivity of the galvanometer is given by,
$V_S={\displaystyle \frac{nBA}{cR}}$…………………….(2)

Where, n is the number of turns in the coil of galvanometer,
B is the magnetic field around the coil
A is the area of the coil
c is the restoring torque per unit twist
R is the resistance of the turn

Complete step by step answer:
First, we can find the current sensitivity of the galvanometer.
The current sensitivity of the galvanometer is given by,
$I_S={\displaystyle \frac{nBA}{c}}$……………………(3)
Where, n is the number of turns in the coil of galvanometer,
B is the magnetic field around the coil
A is the area of the coil
c is the restoring torque per unit twist
It is given that current sensitivity is 5 div/mA.
We can put this value in equation (3).
So, we can write that,
$I_S={\displaystyle \frac{nBA}{c}}=5$
Equation (2) gives the voltage sensitivity of the galvanometer,
$V_S={\displaystyle \frac{nBA}{cR}}$

The voltage sensitivity of the galvanometer is 20div/V.
Now, we will look for a relation between the voltage sensitivity and current sensitivity of the galvanometer.
Combining Equation (1) and (2) will give us,
$V_S={\displaystyle \frac{nBA}{cR}}=20$
So, the resistance of the galvanometer is ,
$V_S={\displaystyle \frac{I_S}{R}}$
$\Rightarrow 20={\displaystyle \frac{{\displaystyle \frac{5}{{10}^{-3}}}}{R}}$
$\Rightarrow R=250$

So, the resistance of the galvanometer is- 250 Ω

Hence the correct option is (A)

Note: As we can see in equation (1), current sensitivity of the galvanometer is independent of the resistance. So, we cannot change the current sensitivity of the galvanometer by changing the resistance of the galvanometer. If you increase the resistance of the galvanometer, it does not increase the sensitivity of the galvanometer. As you have to change the number of turns to increase the resistance, the effect will be nullified, and the voltage sensitivity will not be changed. You need to make sure that you don’t make any unnecessary calculations beforehand. There are lots of unknown variables, and we didn’t care about any of them. Find the final result in terms of given quantities.