Question

In the circuit shown below, if the resistance of voltmeter is $4k\Omega$, then the error in the reading of voltmeter will be

• A. $50$%
• B. $68$%
• C. $17$%
• D. $33.3$%

Answer 1

Answer: (D)

$33.3$%

When the galvanometer is connected, total resistance in the circuit is $4k\Omega$ and $4k\Omega$ in parallel and this combination is in series with $4k\Omega$.
So, total resistance of circuit $R_{eq}=(4\parallel 4)+4=2+4=$ $6k\Omega$.
The current flowing $=\frac{4V}{6000\Omega }=\frac{2}{3}mA$
$P.D.$ across $4k\Omega =4\times 1000\times \frac{2}{3}mA=2.666V$
If the voltmeter had not been connected, equivalent resistance of the circuit $R^{\prime }=4+4=8k\Omega$
The current would have been $\frac{4V}{8000\Omega }=0.5mA$.
The potential difference across $4k\Omega$ would have been $V_{4k\Omega }=4000\Omega \times 0.5mA=2V$
$\therefore$ Error % $=\frac{2.66-2}{2}\times 100=\frac{0.66}{2}\times 100=33.3$%.