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 confi

Question

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.]
NEET 2026 (Re-NEET)

TestHub · Accepted Answer

Let the inductance of each identical inductor be $L$ .

For configuration (P):
The two inductors are connected in series. The points $a$ and $b$ are located across the first inductor only. The current flowing through this inductor is $I (t)$ .
The induced emf between points $a$ and $b$ is given by:
$E_{P} = L \frac{d I}{d t}$

For configuration (Q):
The two inductors are connected in parallel. The points $a$ and $b$ are the common nodes of this parallel combination.
The equivalent inductance $L_{e q}$ between $a$ and $b$ is:
$L_{e q} = \frac{L \times L}{L + L} = \frac{L}{2}$
The total current entering the parallel combination is $I (t)$ . The induced emf between points $a$ and $b$ is:
$E_{Q} = L_{e q} \frac{d I}{d t} = \frac{L}{2} \frac{d I}{d t}$

Now, finding the ratio of the induced emfs:
$\frac{E _{P}}{E _{Q}} = \frac{L \frac{d I}{d t}}{\frac{L}{2} \frac{d I}{d t}} = 2$

Answer: $2$

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