Power in an AC circuit

A heater and a coil on the same socket

A heater turns all the energy it takes into heat. A coil on an AC supply takes energy in one quarter of a cycle and gives it back in the next. So how do we count the power of a circuit that has both kinds of part? We find out what happens to $p = vi$ over one whole cycle.

Power at one instant

Take the voltage $v = V_0\sin\omega t$ and the current $i = I_0\sin(\omega t - \phi)$. The power at one instant is $p = vi$. We open the second sine:

This is two terms. The first is $V_0I_0\sin^2\omega t\,\cos\phi$. The second is $-V_0I_0\sin\omega t\cos\omega t\,\sin\phi$.

Now average over one full cycle. The average of $\sin^2\omega t$ is $\dfrac{1}{2}$. The average of $\sin\omega t\cos\omega t$ is $0$. It equals half of $\sin 2\omega t$, which spends as much time positive as negative. So the average power is

The picture shows a circuit with $\phi = 53^\circ$. For part of every cycle $p$ is negative. In those moments the circuit gives energy back to the source. The average is what is left over.

The power factor

The factor $\cos\phi$ is the power factor. In the impedance triangle, $\cos\phi = \dfrac{R}{Z}$. So

Write $V_{rms} = I_{rms}Z$. The $Z$ cancels:

All the power goes into the resistor. That makes sense. A pure coil and a pure capacitor only store energy and return it.

• A pure resistor. $\phi = 0$, power factor $1$. The power is $V_{rms}I_{rms}$.
• A pure coil or a pure capacitor. $\phi = \pm 90^\circ$, power factor $0$. The average power is zero. The current is called wattless.
• A series LCR circuit. The power factor is between $0$ and $1$.
• At resonance. $\phi = 0$, so the power factor is $1$ and the power is the largest it can be for that source.

The product $V_{rms}I_{rms}$ is the apparent power, in volt-ampere (VA). The real power is $P = V_{rms}I_{rms}\cos\phi$, in watt. The apparent power is the real power only when the power factor is $1$.

A worked example

A circuit has $R = 30\ \Omega$ and a net reactance $X_L - X_C = 40\ \Omega$. It is joined to a source of $200$ V (rms). Find the average power.

1. The impedance. $Z = \sqrt{(30\ \Omega)^2 + (40\ \Omega)^2} = 50\ \Omega$.
2. The current. $I_{rms} = \dfrac{200\ \text{V}}{50\ \Omega} = 4.0$ A.
3. The power factor. $\cos\phi = \dfrac{R}{Z} = \dfrac{30\ \Omega}{50\ \Omega} = 0.6$.
4. The power. $P = V_{rms}I_{rms}\cos\phi = 200\ \text{V} \times 4.0\ \text{A} \times 0.6 = 480$ W.

Check. $I^2R = (4.0\ \text{A})^2 \times 30\ \Omega = 480$ W. It matches. The apparent power is $200 \times 4.0 = 800$ VA, so only $60\ \%$ of it is real power.

Set the net reactance to $0$. Does the curve ever go below zero? Now make $R$ small and the net reactance large. What happens to the average power? Which slider has no effect on the power factor?

Why a low power factor is a nuisance

Suppose a machine must deliver $4600$ W of real power from a $230$ V supply. We want the current, so $I = \dfrac{P}{V\cos\phi}$.

• With a power factor of $1$, the current is $\dfrac{4600\ \text{W}}{230\ \text{V} \times 1} = 20$ A.
• With a power factor of $0.5$, the current is $\dfrac{4600\ \text{W}}{230\ \text{V} \times 0.5} = 40$ A.

The same useful power needs twice the current. The supply wires heat by $I^2r$, so they lose four times as much energy. That is why supply companies want factories to keep the power factor close to $1$. A factory does it by adding capacitors, which cancel the effect of its motors, since a motor behaves like a coil.

A related idea is the choke coil. To limit the current in a tube light we could put a resistor in series, but it would turn energy into heat. A coil of very small resistance cuts the current just as well, because it only has reactance. Its power factor is almost zero, so it uses almost no power.

Practise with new numbers

Try it in Python

The program adds up the power $vi$ at $2000$ equal steps through one cycle, and compares the average with three formulas.

Common mistakes

• Using peak values in $P = VI\cos\phi$. The formula $V_{rms}I_{rms}\cos\phi$ needs rms values. With peak values write $\tfrac{1}{2}V_0I_0\cos\phi$.
• Forgetting $\cos\phi$. $V_{rms}I_{rms}$ alone is the apparent power, in VA.
• Using $P = I^2Z$. Only the resistor uses power, so $P = I^2R$.
• Saying a coil or a capacitor uses power. Alone, each takes zero average power. They store energy and return it.
• Mixing up $\cos\phi$ and $\sin\phi$. The power factor is $R/Z$, not $X/Z$.
• Thinking a capacitive circuit has a negative power. The power factor depends on $\cos\phi$, which is the same for $+\phi$ and $-\phi$.

Remember these

• $p = vi$ goes above and below zero. The average over a cycle is $P = V_{rms}I_{rms}\cos\phi$.
• Power factor: $\cos\phi = \dfrac{R}{Z}$. It is $1$ for a resistor and at resonance, and $0$ for a pure coil or capacitor.
• Only the resistor uses power: $P = I_{rms}^2R$.
• Apparent power $V_{rms}I_{rms}$ is in VA. Real power is in W.
• A low power factor means a larger current for the same power, and more loss in the wires.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An AC circuit takes $2.0$ A (rms) from a $200$ V (rms) supply. The power factor is $0.50$. What is the average power, and what is the apparent power?

2. What is the average power taken by an ideal capacitor on an AC supply? Why?

3. A series circuit is in resonance. What are its power factor and its impedance?

4. A series circuit has $R = 8\ \Omega$ and $Z = 10\ \Omega$. What is its power factor?