AC voltage across an inductor
Why the current lags
Connect an ac source v equals v m sin ωt across a coil of inductance L. Assume the coil’s resistance is negligible: a pure inductor.
The changing current makes a back emf, minus L di by dt. Going round the loop, v minus L di by dt equals zero. So L di by dt equals v m sin ωt.
This says the slope of the current graph follows the voltage. Where the voltage is largest, the current is changing fastest. Where the voltage is zero, the current has stopped changing: it is at a peak.
Integrate: di by dt equals v m over L, sin ωt. The current is i equals minus v m over ωL, cos ωt. The constant of integration is zero, because the current swings evenly about zero.
Using minus cos ωt equals sin of ωt minus π by 2, we get i equals i m sin of ωt minus π by 2, with i m equals v m over ωL.
Plot both. The current reaches its peak a quarter cycle after the voltage. In a pure inductor, the current lags the voltage by π by 2, that is 90°.
On the phasor diagram, the current phasor I is a quarter turn behind the voltage phasor V, and the two turn together.
Compare i m equals v m over ωL with Ohm’s law. The quantity ωL plays the role of resistance. We call it the inductive reactance, X L equals ωL, measured in ohms.
Inductive reactance and power
X L equals ωL, which is 2πνL. So the reactance grows with frequency. Double the frequency, and X L doubles, and the current halves.
For a steady dc current, ν is zero, so X L is zero. An ideal inductor does not oppose a steady current at all. It only opposes changes in current.
Example: a pure inductor of 25 millihenry is connected to a 220 volt, 50 hertz source. X L equals 2π × 50 × 0.025, which is 7.85 ohms.
The rms current is I equals V over X L, that is 220 over 7.85, about 28 amperes.
Now the power. p equals v i, which is v m sin ωt times minus i m cos ωt. That equals minus ½ v m i m sin 2ωt.
Plot p. For a quarter cycle the inductor takes energy from the source and stores it in its magnetic field. In the next quarter it gives it all back.
Positive and negative parts are equal, so the average power over a full cycle is zero. A pure inductor uses no energy, even though a current flows.
A light bulb in series with a coil glows on ac. Push an iron rod into the coil: L rises, X L rises, the current drops, and the bulb grows dimmer. On dc the rod would make no difference.
Try it: ac through an inductor
- This is the example: 25 mH at 220 V, 50 Hz. Check X_L = 7.85 Ω and I ≈ 28 A.
- Watch the graph: the blue current peaks a quarter cycle after the amber voltage. The I phasor trails V by 90°.
- Double the frequency to 100 Hz. X_L doubles and the current halves.
- Tick “Show power”: the green and red areas are equal, so the average power is zero.
Current through a choke
A pure inductor of 0.2 H is connected to a 220 V, 50 Hz supply. What is the rms current?
- 3.50 A
- 4.95 A
- 22 A
- 1100 A
Answer
A. X_L = 2πνL = 2π × 50 × 0.2 = 62.8 Ω. So I = V/X_L = 220/62.8 = 3.50 A. (4.95 A is the peak current, 311/62.8. 22 A comes from using ω = 50 instead of 2π × 50, and 1100 A from dividing V by L alone.)
Energy in an inductor
An rms current I flows through a pure inductor connected to an ac source of rms voltage V. What is the average power taken from the source?
- VI
- ½ VI
- I² X_L
- Zero
Answer
D. p = vi = −½ v_m i_m sin 2ωt, which is positive for one quarter cycle and negative for the next. The inductor stores energy in its magnetic field and then returns all of it to the source, so the average power is zero. VI and I²X_L would be true only if the energy were used up, as in a resistor.
Summary
- Current in a pure inductor:
i = i_m sin(ωt − π/2), i_m = v_m/X_L - Inductive reactance:
X_L = ωL = 2πνL - Average power:
P = 0
- In a pure inductor the current lags the voltage by π/2 (90°), a quarter of a period.
- The inductive reactance X_L limits the current like a resistance. It is measured in ohms and is proportional to frequency; it is zero for dc.
- Energy goes into the magnetic field for a quarter cycle and comes back in the next, so no power is used on average.