Q 12-03-145JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
Figure shows a part of an electric circuit. The potentials at points $a$, $b$ and $c$ are $30\ \text{V}$, $12\ \text{V}$ and $2\ \text{V}$ respectively. The current through the $20\ \Omega$ resistor will be
Answer: (B) $0.4\ \text{A}$
Let the junction be at $x$: $\dfrac{30-x}{10}=\dfrac{x-12}{20}+\dfrac{x-2}{30}$.
Multiply by 60: $180-6x=3x-36+2x-4\Rightarrow x=20\ \text{V}$.
Current in $20\ \Omega$: $\dfrac{20-12}{20}=0.4\ \text{A}$.
Solution by Sreeraj P, M.Sc Physics