AC circuit with inductor

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Homework Statement


Imagine a simple AC circuit with a voltage source and an inductor:
ac_inductor_sch.gif

Determine the current going through the circuit.


Homework Equations


voltage of the source: [tex]v = V_m\sin{\omega}t[/tex]
induced emf in the inductor: [tex]e = L\frac{di}{dt}[/tex]

The Attempt at a Solution


By Kirchhoff's second rule, v - e = 0. What puzzles me is that this would basically mean that there's no current going through the circuit, wouldn't it? But then there would be no induced emf in the inductor, so that's probably not the case. How come there's some current going through the circuit when the induced emf opposes the source emf?
 

Answers and Replies

  • #2
nrqed
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Homework Statement


Imagine a simple AC circuit with a voltage source and an inductor:
ac_inductor_sch.gif

Determine the current going through the circuit.


Homework Equations


voltage of the source: [tex]v = V_m\sin{\omega}t[/tex]
induced emf in the inductor: [tex]e = L\frac{di}{dt}[/tex]

The Attempt at a Solution


By Kirchhoff's second rule, v - e = 0. What puzzles me is that this would basically mean that there's no current going through the circuit, wouldn't it? But then there would be no induced emf in the inductor, so that's probably not the case. How come there's some current going through the circuit when the induced emf opposes the source emf?
the fact that [itex] V_{source} - V_{inductor} = 0 [/itex] does not imply that there is no current!! (think of a simple DC circuit with a single resistor...there again th esum of the potential differences is zero and yet there is a current).

Quite the opposite, for [itex] V_{inductor} [/itex] to be nonzero the current must be changing in time.

So you have to solve the differential equation. But the fastest way to answer the question is to simply use that the impendance of an inductor is given by [itex] \omega L [/itex] and then the current is the voltage of the source divided by the impedance (and is out of phase by 90 degrees)
 

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