Question

The resistance of the coil of a tangent galvanometer is $60\Omega$. It is connected to a battery of negligible internal resistance. The deflection is found to be ${60}^{\circ }$. Now a shunt resistance of $30\Omega$ is connected across the coil of the tangent galvanometer. The deflection $\left(\text{in degree}\right)$ produced will be

Answer 1

Given:
$R=60\Omega ;\theta ={60}^{\circ };S=30\Omega$

Let $V$ be the potential difference across galvanometer.

So initially current flowing through galvanometer will be

$I_i=\frac{V}{R}=\frac{V}{60}$

When shunt is connected to galvanometer, effective resistance become

$R_{eq}=\frac{R\times S}{R+S}=\frac{60\times 30}{60+30}=20\Omega$

So, the new current will be

$I=\frac{V}{20}$


Applying $\text{KVL}$ in above diagram,

$I_{g}R=(I-I_g)S$

$I_g\times 60=(\frac{V}{20}-I_g)\times 30$

$\therefore I_g=\frac{V}{60}$

As we know, that current in the tangent galvanometer is

$i=K\tan \theta$

$i\propto \tan \theta (\text{for}K=\text{constant})$

$\therefore \frac{I_i}{I_g}=\frac{\tan \theta }{\tan {\theta }^{\prime }}$

$\Rightarrow \frac{V/60}{V/60}=\frac{\tan {60}^{\circ }}{\tan {\theta }^{\prime }}$

$\therefore {\theta }^{\prime }={60}^{\circ }$