Question 330 of 380

Two resistance of $100\Omega$ and $200\Omega$ are connected in series with a battery of $4V$ and negligible internal resistance. A voltmeter is used to measure voltage across $100\Omega$ resistance, which gives reading as $1V$. The resistance of voltmeter must be _______ $\Omega$.

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Answer: 200

Voltage across $200\Omega =4-1=3V$.
Therefore, current through the battery, $\dfrac{3}{200}$
Now, equivalent resistance of {R}_{V}&100\Omega,$=\dfrac{{R}_{v}100}{{R}_{v}+100}$
For voltmeter, $\dfrac{{R}_{v}100}{{R}_{v}+100}\times \dfrac{3}{200}=1$
$\Rightarrow 3{R}_{v}=2{R}_{v}+200$
$\Rightarrow {R}_{v}=200\Omega$