Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

Two identical inductors are connected in two different configurations PP and QQ, where a time varying current I(t)I(t) is flowing, as shown in the figure. The induced emf between points aa and bb for configuration PP is EPE_P and that for configuration QQ is EQE_Q. The ratio EP/EQE_P/E_Q is:
[Neglect the effect of mutual inductance.]


[IMAGE]



Show answer and solution
Sol. In figure (P)

EP=Eab=−LdI(t)dtE_P = E_{ab} = -L\frac{dI(t)}{dt} ...(1)

<image_2>


In figure (Q)


EQ=Eab=−L2dI(t)dtE_Q = E_{ab} = -\frac{L}{2}\frac{dI(t)}{dt} ...(2)

<image_3>


⇒EPEQ=21\Rightarrow \frac{E_P}{E_Q} = \frac{2}{1}

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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