Question

When the voltage across a battery is measured using a dc potentiometer, the reading shows 1.08 V. But when the same voltage is measured using a Permanent Magnet Moving Coil (PMMC) voltmeter, the voltmeter reading shows 0.99 V. If the resistance of the voltmeter is $1100\Omega ,$ the internal resistance of the battery, in $\Omega$ is

• A. 100

Answer 1

The correct option is A 100

Open circuit voltage,
$V_S=1.08V$

$R_m=1100\Omega$

$V_0=V_s\frac{R_m}{R_m+R_s}$

$0.99=\left(1.08\right)\frac{\left(1100\right)}{\left(1100\right)+R_s}$

$1089+0.99R_s=1188$

$R_s=\frac{1188-1089}{0.99}=100\Omega$