Q. A $ 36\,\Omega $ galvanometer is shunted by resistance of $ 4\,\Omega . $ The percentage of the total current, which passes through the galvanometer is
Jharkhand CECEJharkhand CECE 2010Current Electricity
Solution:
Let G be resistance of galvanometer and $ {{i}_{g}} $ the current which on passing through the galvanometer produces full scale deflection. If $ i $ is the maximum current, and since, G and S are in parallel.
$ {{i}_{g}}\times G=(i-{{i}_{g}})\times S $
$ \Rightarrow $ $ \frac{{{i}_{g}}}{i}=\frac{S}{S+G} $
Given, $ G=36\,\Omega ,\,S=4\Omega $
$ \therefore $ $ \frac{{{i}_{g}}}{i}=\frac{4}{36+4}=\frac{4}{40} $
$ \Rightarrow $ $ {{i}_{g}}=\frac{i}{10} $
$ \therefore $ $ 1000\times \frac{{{i}_{g}}}{i}=\frac{1}{10}\times 100 $
$ \Rightarrow $ $ \frac{{{i}_{g}}}{i}%=10% $
