Current Electricity Question 21
Question 21 - 2024 (30 Jan Shift 2)
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 $\Omega$.
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Answer: (200)
Solution:
$$ \begin{aligned} & \frac{R_{v} 100}{R_{v}+100}=\frac{200}{3} \ & 3 R_{v}=2 R_{v}+200 \ & R_{v}=200 \end{aligned} $$
