Question

A galvanometer of resistance $95\Omega$ shunted by a resistance of $5\Omega$ gives a deflection of $50$ divisions when joined in series with a resistance of $20\text{k}\Omega$ and a $2\text{V}$ accumulator. What is the current sensitivity of the galvanometer in $\text{div}/\mu \text{A}$?

• A. $10$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $10$

As $20\text{k}\Omega$ is much greater than the resistance of the shunted galvanometer $5\Omega$,the current in the circuit will be

$I=\frac{2}{20\times {10}^3}={10}^{-4}\text{A}=100\mu \text{A}$
Now, let us say, $I_g$ is the current passing through galvanometer, then, $(I-I_g)$ current will flow through $5\Omega$ resistor.
Since, both are parallel, the potential drop across both the elements will be equal.
$\Rightarrow 95I_g=5(I-I_g)\Rightarrow I_g=\frac{I}{20}=5\mu A$
And as this current produces a deflection of $50$ divisions in the galvanometer, so

Current sensitivity,

$\text{CS}=\frac{\theta }{I_g}=\frac{50\text{div}}{5\mu \text{A}}=10\frac{\text{div}}{\mu \text{A}}$

Hence, option (a) is the correct answer.

Key concept: Deflection per unit current is called current sensitivity (C.S) of the galvanometer $$\text{CS}=\frac{\text{Deflection}\left(\theta \right)}{\text{Current}\left(I\right)}$$Note : Shunting a galvanometer decreases its current sensitivity.