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) $2$
Let each inductor have inductance $L$.
Configuration P: points $a$ and $b$ are across one inductor, which carries the full current $I(t)$:
$$E_P = L\frac{dI}{dt}$$
Configuration Q: the two inductors are in parallel between $a$ and $b$, so each carries $I/2$:
$$E_Q = L\frac{d(I/2)}{dt} = \frac{L}{2}\frac{dI}{dt}$$
(Equivalently, the parallel combination has inductance $L/2$.)
$$\frac{E_P}{E_Q} = 2$$
Solution by Sreeraj P, M.Sc Physics