Two identical inductors are connected in two different configurations P and Q, where a time varying current $I(t)$ is flowing, as shown in the figure. The induced emf between points $a$ and $b$ for configuration P is $E_P$ and that for configuration Q is $E_Q$. The ratio $E_P/E_Q$ is :
[Neglect the effect of mutual inductance.]
Answer: (D) $2$
Concept. The emf across an inductor is $L\dfrac{dI}{dt}$, where $I$ is the current through that inductor.
Step 1. Configuration P. $a$ and $b$ are across one inductor carrying the full current: $E_P = L\dfrac{dI}{dt}$.
Step 2. Configuration Q. The two inductors are in parallel between $a$ and $b$, each carrying $\dfrac{I}{2}$: $E_Q = \dfrac{L}{2}\dfrac{dI}{dt}$.
Step 3. Ratio. $\dfrac{E_P}{E_Q} = 2$.
Answer. $2$ (Option D).
Solution by Sreeraj P, M.Sc Physics