Q 12-03-242JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
In the given circuit the internal resistance of the $18\ \text{V}$ cell is negligible. If $R_1 = 400\ \Omega$, $R_3 = 100\ \Omega$ and $R_4 = 500\ \Omega$ and the reading of an ideal voltmeter across $R_4$ is $5\ \text{V}$, then the value of $R_2$ will be:
Answer: (B) $300\ \Omega$
Current in the $R_3$-$R_4$ branch: $I_1 = \dfrac{5}{500} = 0.01\ \text{A}$, so the voltage across that branch (and across $R_2$) is $0.01\times600 = 6\ \text{V}$.
The remaining $18-6 = 12\ \text{V}$ is across $R_1$, so the total current is $\dfrac{12}{400} = 0.03\ \text{A}$.
Current in $R_2$: $0.03-0.01 = 0.02\ \text{A}$, so
$$R_2 = \frac{6}{0.02} = 300\ \Omega$$
Solution by Sreeraj P, M.Sc Physics