Question

The resistance of a voltmeter is $100\,\Omega$ . It is connected to a $12\,V - 20\,W$ battery. The reading of voltmeter is (Nearly)
(A) $6\,V$
(B) $8\,V$
(C) $9\,V$
(D) $11\,V$

Answer 1

From the power formula given below, calculate the value of the internal resistance by substituting the known parameters in it. Then substitute the internal resistance in the ohm’s law to get current. Multiply the resistance with the current to get the reading of voltmeter.

Useful formula:
(1) The power is given by
$P = \dfrac{{{V^2}}}{r}$
Where $P$ is the power of the battery, $V$ is the potential difference developed and $r$ is the internal resistance of the voltmeter.
(2) The ohm’s law is given by
$V = IR$
Where $I$ is the current flowing through the voltmeter.

Complete step by step solution:
It is given that the
Resistance of the voltmeter, $R = 100\,\Omega$
Potential difference developed in the voltmeter, $V = 12\,V$
The power of the battery, $P = 20\,W$
Use the formula (1) ,
$P = \dfrac{{{V^2}}}{r}$
Rearranging the given equation, we get
$r = \dfrac{{{V^2}}}{P}$
Substituting the values of the potential difference and the internal resistance in the above equation,
$r = \dfrac{{{{12}^2}}}{{20}} = \dfrac{{144}}{{20}}$
By performing the basic arithmetic operations,
$r = 7.2\,\Omega$
Hence the internal resistance is obtained as $7.2\,\Omega$ . Using the formula (2) of the ohm’s law,
$V = IR$
The resistance here is the sum of the both internal resistance and the resistance of the voltmeter.
$V = I\left( {R + r} \right)$
By rearranging the equation to find the current value,
$I = \dfrac{V}{{\left( {R + r} \right)}}$
$I = \dfrac{{12}}{{\left( {100 + 7.2} \right)}}$
By performing the basic arithmetic operation, we get

$I = 0.1119A$
In order to obtain the reading of the voltmeter, multiply the resistance with the current,
Reading = $0.1119 \times 100 = 11.19 \approx 11\,V$
Hence the voltmeter reading is obtained as $11\,V$ .

Thus the option (D) is correct.

Note: The internal resistance, the term specified in voltmeter is the opposition to the current flow by its own cell and the batteries itself. Hence they cause generation of the heat to the surroundings. The reason for this is sulfation and the grid corrosion of the batteries.