Q 12-06-065JEE MainJEE Main 2023 (30 Jan, Shift 1)Medium
As per the given figure, if $\dfrac{dI}{dt}=-1\ \text{A s}^{-1}$, then the value of $V_{AB}$ at this instant will be ______ V.
Numerical value type. Enter your answer.
Answer: 30
Let $P$ be the left corner. The current $2\ \text{A}$ flows from $A$ through $R$ and $L$ towards $P$, so
$$V_A-V_P=IR+L\frac{dI}{dt}=2\times12+6\times(-1)=18\ \text{V}$$
The cell's positive terminal faces $P$, so $V_P-V_B=12\ \text{V}$.
$$V_{AB}=V_A-V_B=18+12=30\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics