Lesson 4 of 10 · 5 min
AC through an inductor
NCERT §7.4
Riya puts her 25.0 mH coil, with negligible resistance, straight across the 220 V, 50 Hz supply. A dc supply would drive a huge current through it. What does ac do?
The lesson in notes
In short
For a pure inductor (negligible resistance), the loop rule gives v − L di/dt = 0, so di/dt = (vm/L) sin ωt.
Integrating, i = −(vm/ωL) cos ωt = im sin(ωt − π/2), with im = vm/ωL. There is no constant term, because the source swings evenly about zero.
The quantity ωL plays the part of resistance. It is the inductive reactance XL = ωL, measured in ohm, and im = vm/XL.
XL grows in proportion to both the inductance and the frequency.
The current lags the voltage by π/2, a quarter of a cycle: it reaches each maximum a quarter period after the voltage does.
The average power into a pure inductor over a cycle is zero. Energy stored in its field during one quarter cycle is handed back to the source in the next.
Example 7.2: a pure 25.0 mH inductor on 220 V, 50 Hz has XL = 2π × 50 × 25 × 10⁻³ = 7.85 Ω and an rms current of 220/7.85 = 28 A.