Question

A galvanometer shows a reading of $0.65\text{mA}$. When a galvanometer is shunted with a $4\Omega$ resistance, the deflection is reduced to $0.13\text{mA}$. If the galvanometer is further shunted with a $2\Omega$ wire, the new reading will be (the main current remains the same)

• A. $0.60\text{mA}$
• B. $0.08\text{mA}$
• C. $0.12\text{mA}$
• D. $0.05\text{mA}$

Answer 1

The correct option is D $0.05\text{mA}$

Given $I=0.65\text{mA}$
When shunted with resistance of $4\Omega$, galvanometer reading is $I^{\prime }=0.13\text{mA}$
$I^{\prime }=\frac{4}{4+G}I\Rightarrow \frac{4}{4+G}=\frac{1}{5}$
$\Rightarrow G=16\Omega$

When galvanometer is further shunted with $2\Omega$ resistance, $R_{eq}=\frac{2\times 4}{2+4}=\frac{4}{3}$
let galvanometer reading be $I^{\prime \prime }$


$I^{\prime \prime }=\frac{4/3}{\frac{4}{3}+16}I$
$I"=\frac{1}{13}I=0.05\text{mA}$