Question

When a galvanometer is shunted with a $4$ Ω resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a $2$ Ω wire, determine current in galvanometer now if initially current in galvanometer is $I_0$. (Given: Main current remains same)

• A. $\frac{I_0}{13}$
• B. $\frac{I_0}{5}$
• C. $\frac{I_0}{8}$
• D. $\frac{5I_0}{13}$

Answer 1

The correct option is A

$\frac{I_0}{13}$

In parallel, current distributes in inverse ratio of resistance

In figure $\left(ii\right)$
$\frac{4I_0/5}{I_0/5}=\frac{G}{4}\therefore G=16\Omega$

In figure $\left(iii\right)$
combined resistance of $4\Omega$ and $2\Omega$ is $4/3\Omega$

Assuming $x{I}_0$ current passes through galvanometer, then $(1-x)I_0$ will be the current passing through $4/3\Omega$.
Hence we can write:
$\frac{x{I}_0}{(1-x)I_0}=\frac{4/3}{16}$
$\frac{x{I}_0}{(1-x)I_0}=\frac{1}{12}$
$12x{I}_0=I_0-x{I}_0$
$13x{I}_0=I_0$
$x=\frac{1}{13}$

So, the current in galvanometer is $x{I}_0=\frac{I_0}{13}$