AC voltage across a capacitor

Why the current leads

Narrated video. Transcript:

Connect a capacitor to a battery. Current flows for a moment while it charges, then stops. In a dc circuit, a capacitor blocks the current.

Now connect it to an ac source. The capacitor charges, discharges, and charges the other way, again and again. Charge keeps flowing to and fro in the wires, so a current flows all the time.

The source voltage v equals v m sin ωt appears across the plates. The charge on the plates is q equals C v, so q equals C v m sin ωt.

The current is the rate of flow of charge, i equals dq by dt. Differentiating, i equals ω C v m cos ωt.

Using cos ωt equals sin of ωt plus π by 2, this is i equals i m sin of ωt plus π by 2, where i m equals ω C v m.

Look at the graphs. The current is largest when the voltage is zero and the plates are charging fastest. It is zero when the voltage is at its peak and the capacitor is full.

So the current reaches its peak a quarter cycle before the voltage. In a capacitor, the current leads the voltage by π by 2, that is 90°. The I phasor is a quarter turn ahead of V.

Write i m as v m divided by 1 over ωC. Then 1 over ωC acts like a resistance. It is the capacitive reactance, X C equals 1 over ωC, measured in ohms.

Capacitive reactance and power

Narrated video. Transcript:

X C equals 1 over 2πνC. Unlike an inductor’s, this reactance falls as the frequency rises. Double the frequency, and X C halves, so the current doubles.

For dc, ν is zero and X C becomes infinite. That matches what we saw: a capacitor blocks a steady current.

Example: a 15 microfarad capacitor is connected to a 220 volt, 50 hertz source. X C equals 1 over 2π × 50 × 15 × 10⁻⁶, which is about 212 ohms.

The rms current is I equals V over X C, 220 over 212, which is 1.04 amperes. The peak current is √2 × 1.04, about 1.47 amperes.

If the frequency is doubled to 100 hertz, X C falls to 106 ohms and the current doubles to about 2.07 amperes.

Power: p equals v i, which is v m sin ωt times i m cos ωt, or ½ v m i m sin 2ωt. Over one cycle its average is zero, just as for the inductor.

The capacitor stores energy in its electric field while it charges, and gives it back while it discharges. No energy is used on average.

A bulb in series with a capacitor: on dc it does not glow. On ac it glows. Reduce the capacitance, and X C rises, so the current falls and the bulb grows dimmer.

Try it: ac through a capacitor

Interactive simulation. Things to try:

• This is the example: 15 μF at 220 V, 50 Hz. Check X_C ≈ 212 Ω and I ≈ 1.04 A.
• The blue current peaks a quarter cycle before the amber voltage. The I phasor is 90° ahead of V.
• Double the frequency: X_C halves and the current doubles. Now try 5 Hz: almost no current flows.
• Tick “Show power”. Over a full cycle the energy given back equals the energy taken.

Current through a capacitor

A 100 μF capacitor is connected to a 220 V, 50 Hz supply. What is the rms current?

1. 0 A: no charge crosses the gap between the plates
2. 6.91 A
3. 9.77 A
4. About 7000 A

Bulb and capacitor

A bulb and a capacitor are connected in series to an ac source. The frequency of the source is increased. What happens to the bulb?

1. It glows more brightly
2. It grows dimmer
3. No change: frequency does not affect a capacitor
4. It goes out, as a capacitor blocks ac

Summary

• Current through a capacitor: i = i_m sin(ωt + π/2), i_m = v_m/X_C
• Capacitive reactance: X_C = 1/ωC = 1/(2πνC)
• Average power: P = 0

• In a capacitor the current leads the voltage by π/2 (90°).
• X_C is measured in ohms and is inversely proportional to frequency. For dc it is infinite, so a capacitor blocks dc but lets ac through.
• Energy is stored in the electric field and returned each cycle, so a pure capacitor uses no power on average.
• Compare: X_L rises with frequency, X_C falls with it.