Power in AC circuits and power factor

Average power and the power factor

Narrated video. Transcript:

In a general ac circuit the source voltage is v equals v m sin ωt, and the current is i equals i m sin of ωt plus φ. How much power does the source deliver on average?

The instantaneous power is p equals v i. Using a product-to-sum formula, p equals v m i m over 2, times cos φ minus cos of 2ωt plus φ.

Plot p. It has a steady part, v m i m over 2 cos φ, and a part that wiggles at twice the source frequency. The wiggling part averages to zero over a cycle.

So the average power is P equals v m i m over 2, cos φ. Writing v m equals √2 V and i m equals √2 I, this becomes P equals V I cos φ.

Here V and I are rms values, and cos φ is called the power factor. For a series LCR circuit, the impedance triangle shows that cos φ equals R over Z.

Case one, a pure resistor: φ is zero, cos φ is 1, and P equals V I. All the power is used.

Case two, a pure inductor or a pure capacitor: φ is plus or minus 90°, cos φ is zero, and no power is used, even though a current flows. Such a current is called a wattless current.

Case three, a series LCR circuit: P equals V I cos φ, and since cos φ equals R over Z and V equals I Z, P equals I² R. Power is dissipated only in the resistor.

Case four, an LCR circuit at resonance: X L equals X C, φ is zero, and the power factor is 1. The circuit takes the maximum possible power, V² over R.

Power factor in practice

Narrated video. Transcript:

Example: a voltage of peak value 283 volts and frequency 50 hertz is applied to a series LCR circuit with R equals 3 ohms, L equals 25.48 millihenry and C equals 796 microfarad.

X L equals 2π × 50 × 25.48 × 10⁻³, which is 8 ohms. X C equals 1 over 2π × 50 × 796 × 10⁻⁶, which is 4 ohms.

So Z equals √(3² plus 4 minus 8, squared), which is √25, or 5 ohms. tan φ equals X C minus X L over R, minus 4 over 3. So φ is minus 53.1°: the current lags.

The current amplitude is 283 over 5, which is 56.6 amperes, so the rms current is 56.6 over √2, which is 40 amperes.

The power dissipated is I² R, 40² × 3, which is 4800 watts. The power factor is cos 53.1°, which is 0.6.

Now tune the source to resonance. ω₀ equals 1 over √(LC), which is 222 radians per second, about 35.3 hertz. Then Z equals 3 ohms, I is 66.7 amperes and the power rises to about 13.3 kilowatts.

Why does the power factor matter to an electricity company? A factory needs a fixed power P at voltage V. The current it draws is I equals P over V cos φ.

Take 10 kilowatts at 220 volts. With cos φ equal to 1, I is 45.5 amperes. With cos φ equal to 0.5, I doubles to 91 amperes, and the I² R heating in the supply lines becomes four times as large.

Motors are inductive, so their current lags. We split the current into I cos φ, in phase with the voltage, and the wattless part I sin φ.

A capacitor connected in parallel draws a leading current that cancels much of the wattless part. The power factor moves towards 1, and the lines carry less current for the same useful power.

Try it: power in an LCR circuit

Interactive simulation. Things to try:

• This is the example: X_L = 8 Ω, X_C = 4 Ω, Z = 5 Ω. Check I ≈ 40 A, cos φ = 0.6 and P ≈ 4.8 kW.
• In the power graph, the red areas are energy flowing back to the source. The dashed average line is P = VI cos φ.
• Press “Go to resonance”: the red areas vanish, cos φ = 1 and the power rises to about 13.3 kW.
• Switch to “L only” or “C only”: the green and red areas become equal and the average power is zero.

Average power

An ac circuit draws an rms current of 10 A from a 220 V supply, with a power factor of 0.5. What average power does it take?

1. 2200 W
2. 1905 W
3. 1100 W
4. 550 W

Power factor of an LCR circuit

A series LCR circuit has R = 3 Ω, X_L = 8 Ω and X_C = 4 Ω. What is its power factor?

1. 0.6
2. 0.75
3. 0.8
4. 1, because the circuit is in series

Summary

• Average power: P = VI cos φ (V, I rms)
• Power factor (series LCR): cos φ = R/Z
• Series LCR: P = I²R

• Only the part of the current in phase with the voltage, I cos φ, delivers power. The part I sin φ is a wattless current.
• Resistor: cos φ = 1. Pure L or C: cos φ = 0 and P = 0. At resonance cos φ = 1 and the power is largest.
• A low power factor needs a larger current for the same power, so more energy is wasted as I²R in the supply lines.
• The power factor of inductive loads is improved by connecting a capacitor in parallel.