Question

A voltmeter having a resistance of 1800 $\Omega$ is employed to measure the potential difference across $200\Omega$ resistance, which is connected to dc power supply of 50 V and internal resistance $20\Omega$. What is the approximate percentage change in the potential difference across $200\Omega$ resistance as a result of connecting the voltmeter across it?

• A. $2.2\%$
• B. $5\%$
• C. $10\%$
• D. $20\%$

Answer 1

Answer: (A)

$2.2\%$

The voltage $V_1$ in the circuit before connecting the voltmeter is
$V_1=E-ir$
$V_1=50-\frac{50}{220}\times 20$
$V_1=50-4.6=45.4V$
When the voltmeter is connected in the circuit as shown in figure, the voltage $V_2$ is
$V_2=50-\frac{50}{1800}\times 200$
$V_2=44.4V$
Now, percentage change in voltages is $=\frac{V_1-V_2}{V_1}\times 100=2.2\%$