Question

A galvanometer coil has a resistance $90\Omega$ and full scale deflection current $10\text{mA}$. A $910\Omega$ resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.1\text{V}$, the number of divisions on its scale is

• A. $90$
• B. $91$
• C. $100$
• D. None

Answer 1

The correct option is C $100$

Given:
$G=90\Omega ;I_g=10\text{mA};S=910\Omega$


Applying formula for conversion of galvanometer into voltmeter

$V=I_g(G+S)$

To give maximum reading , full scale deflection current should pass through the galvanometer.

$V=(10\times {10}^{-3})(90+910)=10\text{V}$

So, number of division $n$ will be

$n=\frac{V}{\text{Least count}}=\frac{10}{0.1}=100$

Hence, option (c) is the correct answer.