Question

The range of a voltmeter is $10V$ and its internal resistance is $50\Omega$. To convert it to a voltmeter of range $15V$, how much resistance is to be added?

• A. Add $25\Omega$ resistor in parallel
• B. Add $25\Omega$ resistor in series
• C. Add $125\Omega$ resistor in parallel
• D. Add $125\Omega$ resistor in series

Answer 1

Answer: (B)

Add $25\Omega$ resistor in series

Voltmeter is actually a galvanometer in series with a large resistance. At maximum voltage range, current is maximum. The resistance of galvanometer can be ignored.

Hence maximum current is given by $I=\frac{10}{50}=0.2A$

New voltmeter range is $15V$. Maximum current does not change as galvanometer is in series and same current flows through it.

Hence, new internal resistance of voltmeter is $R=\frac{V}{I}=\frac{15}{0.2}=75\Omega$

Two resistances are in series, $R_{eq}=R_1+R_2$

So, $75=50+R_2$

or, $R_2=25\Omega$