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Question
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Solution
The diagram below depicts an alternating current source generating a voltage e = e0 sin wt and connecting it to a key K and a pure inductor of inductance L to form a closed circuit.
An AC source connected to an inductor
When the key K is closed, an emf is induced in the inductor as the magnetic flux associated with it changes over time. This emf opposes the applied emf, and we have, according to Faraday and Lenz's laws of electromagnetic induction,
`"e'" = - "L""di"/"dt"` ....(1)
The induced emf is e', while the current through the inductor is i. To keep the current flowing, e and e' must have the same magnitude but opposing directions.
According to Kirchhoff's voltage law, as the resistance of the inductor is assumed to be zero, we have, e = -e' = L`"di"/"dt"` ....(2)
∴ `"di"/"dt" = "e"/"L" = ("e"_0 sin omega"t")/"L"`
∴ `int "di" = int ("e"_0 sin omega"t")/"L" "dt"`
∴ i = `- "e"_0/(omega"L") cos omega"t" + "C"`
where C is the integration constant. C must be time-independent and have a current dimension. Because e oscillates around zero, I must similarly oscillate around zero, and so no time independent component of current can exist.
∴ C = 0
∴ i = - `"e"_0/(omega "L") cos omega"t" = - "e"_0/(omega"L") sin (pi/2 - omega"t")`
∴ i = `"e"_0/(omega"L") sin (omega"t" - pi/2)` .....(3)
as sin (- θ) = - sin θ
From Eq. (3), `"i"_"peak" = "i"_0 = "e"_0/(omega "L")`
∴ i = `"i"_0 sin (omega"t" - pi/2)` .....(4)
When this equation is compared to e = e0 sin wt, it is seen that e leads i by π/2 rad, indicating that the voltage is π/2 rad ahead of the current in phase.
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