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When we calculate open circuit voltages for a non ideal transformer . We have current I2 in the right side loop =0 hence the current passing through L3 must be zero but as visible there exists an applied voltage across the inductor ! (If we assume the outer loop on left side and apply Kirchoff's Voltage Law ) How is this possible ?

Is it possible for an inductor to have a voltage across it without any current passing through it ?

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  • $\begingroup$ "We have current I2 in the right side loop =0 hence the current passing through L3 must be zero" How do you mean? Which element is experiencing current I2? $\endgroup$ Commented Apr 5, 2015 at 12:59
  • $\begingroup$ the inductor L4 , if i calculate open circuit voltage . $\endgroup$
    – v_g
    Commented Apr 5, 2015 at 13:05

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Is it possible for an inductor to have a voltage across it without any current passing through it ?

Yes, if electromotive force due to core counteracts the applied voltage by the rest of the network. For harmonic driving voltage the current oscillates harmonically so this occurs only at one instant.

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If you take a simple circuit of battery-switch-inductor, and close the switch, at the moment of closing the inductor sees a voltage but no current, because the current has to build up over time, as energy is put into the magnetic field. If instead of a battery you have an AC source and no switch, voltage and current across the inductor are out of phase, so there are points when the current is zero and the voltage is not. A circuit can have impedence, measured in ohms, even without power dissipation.

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