Question

We have a galvanometer of resistance $25\Omega$. It is shunted by $2.5\Omega$ wire. The part of the total current that flows through the galvanometer is given as :

• A. $\frac{i}{i_0}=\frac{4}{11}$
• B. $\frac{i}{i_0}=\frac{3}{11}$
• C. $\frac{i}{i_0}=\frac{2}{10}$
• D. $\frac{i}{i_0}=\frac{1}{11}$

Answer 1

The correct option is D $\frac{i}{i_0}=\frac{1}{11}$

When a galvanometer is connected to the shunt resistance, then we have

potential drop across the galvanometer $=$ potential drop across the shunt

i.e. $iG=(i_0-i)S$ ...(i)

Here, $i=$ current through the galvanometer

$G=$ resistance of the galvanometer

$i_0=$ total current

$S=$ shunt resistance

From equation (i), we have the part of the total current that flows through the galvanometer is

$\frac{i}{i_0}=\frac{S}{G+S}$

Substituting, $S=2.5\Omega ,G=25\Omega$, we get

$\frac{i}{i_0}=\frac{1}{11}$