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At, $t=0$, you just start moving the coil of the A.C source, in a magnetic field, such that an emf is induced.

Initially, $E=0$, and what I believe is that the current is just flowing in the circuit through the inductor. But what I've read in the textbook is that the current is already in its peak value at $t=0$, when $E=0$ where its phasor lags behind the voltage's phasor by $90^o$.

How is this even possible?

Please provide Current-Emf-time graph starting from $t=0$ as well.

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That is only true if the inductor was previously magnetized. If you aren't moving the coil prior to the moment of t=0, the current Will start increasing at the same time as voltage, but the current Will be smaller than in the conductor.

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  • $\begingroup$ The current must be same though out the circuit. How can it get smaller? More explanation perhaps? $\endgroup$ – Deechit Poudel Mar 3 '17 at 12:22
  • $\begingroup$ It should be easier to understand if you had an AC voltage source and a switch. Before you turn on the switch the current can't flow, so it's 0. Once you turn the switch, the current Will start flowing the same time as voltage and Will be increasing at the same time. However, once the voltage reaches the maximum value, the current Will still be increasing, even though the voltage is falling. That current Will come from the magnetic field created by the coil.This is where the phase shift is created between the current and the voltage. It operates that way from here on. $\endgroup$ – MaDrung Mar 6 '17 at 6:51

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