Question

The range of voltmeter is 10 V and its internal resistance is $50\Omega$. To convert it to a voltmeter of range 15 V, how much resistance is to be added?
(A) Add $25\Omega$resistor is parallel
(B) Add $25\Omega$ resistor is series
(C) Add $125\Omega$ resistor in parallel
(D) Add $125\Omega$ resistor in series

Answer 1

The range of voltmeter can be increased by connecting a suitable high resistor in series with it. In order to do this first, we need to calculate resistance per volt which is given by
$r = \dfrac{{{R_{given}}}}{{{V_{given}}}}$
Where
${R_{given}}$ is internal resistance
${V_{given}}$ is the initial range of the voltmeter.
Since the range of the voltmeter is increased to 15 V therefore we should also increase the total effective resistance across the voltmeter by connecting a suitable resistance in series with it.

Complete step by step solution:
According to the initial data of the question we have
Initial Range of voltmeter is 10 V
New range of voltmeter is 15 V
Resistance per volt, r given by
$r = \dfrac{{{R_{given}}}}{{{V_{given}}}} \Leftrightarrow r = \dfrac{{50}}{{10}}$
$r = 5\Omega$
Therefore total effective resistance of new voltage
$R = 15 \times r$
$R = 15 \times 5 = 75\Omega$
Total Resistance we need $= R - {R_{given}}$
$R - {R_{given}}$ $= 75 - 50$
On further solving we get
$R - {R_{given}}$ $= 25\Omega$
So, we should add a $25\Omega$ resistor in series in order to convert $10{\text{V}}$voltmeter into $15{\text{V}}$ voltmeter.

Hence, Option (b) is correct.

Note: The range of a voltmeter can decrease by reducing its resistance. This can be done by putting a suitable resistance in parallel with the voltmeter. The voltmeters are designed to give full-scale deflection when a certain amount of current passes through it. Series connected resistor will determine the range of the voltmeter and the scale is accordingly marked. Similarly, a parallel-connected shunt with determining its current range.