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 $4 V$ and negligible internal resistance. A voltmeter is used to measure voltage across $100 \Omega$ resistance, which gives reading as $1 V$. The resistance of voltmeter must be $\Omega$.
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Answer: (200)
Solution:
$$ \begin{aligned} & \frac{R _v 100}{R _v+100}=\frac{200}{3} \\ & 3 R _v=2 R _v+200 \\ & R _v=200 \end{aligned} $$
