Q 12-03-106JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
Two resistances of $100\ \Omega$ and $200\ \Omega$ are connected in series with a battery of $4\ \text{V}$ and negligible internal resistance. A voltmeter is used to measure the voltage across the $100\ \Omega$ resistance, and it gives a reading of $1\ \text{V}$. The resistance of the voltmeter must be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 200
The $200\ \Omega$ resistor then has $3\ \text{V}$, so the current is $I = 3/200 = 15\ \text{mA}$.
This current flows through the $100\ \Omega$ resistor and voltmeter in parallel, with $1\ \text{V}$ across them:
$$\frac1{100} + \frac1{R_V} = \frac{15\times10^{-3}}{1} \Rightarrow \frac1{R_V} = 0.005 \Rightarrow R_V = 200\ \Omega$$
Solution by Sreeraj P, M.Sc Physics