Question 402: Current Electricity

Question

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 _______ $$\Omega$$.

Accepted Answer

$$\begin{aligned}
& \frac{R_v 100}{R_v+100}=\frac{200}{3} \
& 3 R_v=2 R_v+200 \
& R_v=200
\end{aligned}$$

Current Electricity

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

Choose the correct answer: