JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 26)
Two resistance of $$100 \Omega$$ and $$200 \Omega$$ are connected in series with a battery of $$4 \mathrm{~V}$$ and negligible internal resistance. A voltmeter is used to measure voltage across $$100 \Omega$$ resistance, which gives reading as $$1 \mathrm{~V}$$. The resistance of voltmeter must be _______ $$\Omega$$.
Answer
200
Explanation
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$$\begin{aligned} & \frac{R_v 100}{R_v+100}=\frac{200}{3} \\ & 3 R_v=2 R_v+200 \\ & R_v=200 \end{aligned}$$
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