Alternating voltage and rms values

A bulb that flickers

A bulb on the mains is not as steady as it looks. Its light dims and brightens again and again, too fast for your eye to see. The reason is the voltage of the socket. It does not stay at one value. It swings from positive to negative and back, fifty times every second in most countries, and sixty times in a few.

A voltage like this is an alternating voltage, or AC. A battery gives a direct voltage, or DC, which keeps one sign. Let us find out how to describe AC, and how to say how big it is.

The sine wave

The voltage of the mains follows a sine curve:

Here $v$ is the instantaneous voltage, its value at the time $t$. The largest value is $V_m$, the peak value. The number $\omega$ is the angular frequency, in rad/s.

• The period $T$ is the time for one full cycle.
• The frequency $f = 1/T$ is the number of cycles in each second, in hertz (Hz).
• One cycle is $2\pi$ rad, so $\omega = 2\pi f$.

For the mains at $f = 50$ Hz, the period is $T = 20$ ms and $\omega = 100\pi = 314$ rad/s. The light of a bulb is brightest at every peak, positive or negative. So it brightens $100$ times each second, not $50$.

A worked example. An AC voltage is $v = 84 \sin(120\pi t)$, with $v$ in V and $t$ in s. Find the peak value, the frequency, the period and the voltage at $t = 2.0$ ms.

1. The peak value. Compare with $v = V_m \sin \omega t$. So $V_m = 84$ V.
2. The frequency. Here $\omega = 120\pi\ \text{rad/s} = 377\ \text{rad/s}$. We want $f$, so $f = \dfrac{\omega}{2\pi} = 60$ Hz.
3. The period. $T = \dfrac{1}{f} = \dfrac{1}{60\ \text{Hz}} = 16.7$ ms.
4. The voltage at $t = 2.0$ ms. The angle is $\omega t = 120\pi \times 0.0020 = 0.754$ rad. Set the calculator to radians. Then $v = 84\ \text{V} \times \sin(0.754) = 84\ \text{V} \times 0.684 = 57.5$ V.

Choose $V_m = 325$ V. Watch the dot. Where is the voltage changing fastest, and where is it changing least?

An AC voltage across a resistor

Join a resistor $R$ to the source. At every instant the voltage across the resistor is $v$, and Ohm's law holds at every instant:

So the peak current is $I_m = V_m/R$. The current rises, falls and changes direction exactly together with the voltage. We say that the current is in phase with the voltage.

What about the power? At each instant $p = i^2 R = I_m^2 R \sin^2 \omega t$. A square is never negative. When the current flows backwards it heats the resistor just as much as when it flows forwards. So a resistor gets hot on AC, even though the average of the current is zero.

The average of $\sin^2 \omega t$ over a cycle is $\tfrac{1}{2}$, because $\sin^2 \omega t = \tfrac{1}{2}(1 - \cos 2\omega t)$ and the cosine averages to zero. So the average power is

The rms value

The average of an alternating current over a cycle is zero. It tells us nothing about the heating. So we use another number. The rms value $I_{rms}$ is the steady direct current that gives the same average power in a resistor:

Set this equal to $\tfrac{1}{2} I_m^2 R$. Then

The letters mean root, mean, square. To find the rms value, square the current at every instant, find the mean of those squares, and take the root.

• An AC voltmeter or ammeter reads the rms value. The rating of a lamp or a socket is an rms value.
• The mains of $230$ V rms has a peak value of $230\ \text{V} \times \sqrt{2} = 325$ V. The insulation of a cable must stand the peak.
• With rms values, the power formulas look just like those of direct current: $P = V_{rms} I_{rms} = I_{rms}^2 R = \dfrac{V_{rms}^2}{R}$.

The mean over half a cycle. Over a full cycle the mean is zero. Over one half cycle, from one zero to the next, the mean of a sine wave is $\dfrac{2}{\pi}\, I_m = 0.637\, I_m$. Some meters and rectifiers work with this number. It is not the rms value. The ratio of the rms value to this mean is $1.11$.

A worked example. The source of the first example, with $V_m = 84$ V and $f = 60$ Hz, is joined to a resistor of $46\ \Omega$. Find the rms current, the average power and the greatest power.

1. The rms voltage. $V_{rms} = \dfrac{84\ \text{V}}{\sqrt{2}} = 59.4$ V.
2. The rms current. $I_{rms} = \dfrac{V_{rms}}{R} = \dfrac{59.4\ \text{V}}{46\ \Omega} = 1.291$ A.
3. The average power. $P = V_{rms} I_{rms} = 59.4\ \text{V} \times 1.291\ \text{A} = 76.7$ W.
4. The greatest power. At the peak, $p = V_m I_m = 84\ \text{V} \times \dfrac{84\ \text{V}}{46\ \Omega} = 153.4$ W. This is twice the average, as it must be.

What if the source were direct? A steady $59.4$ V on the same resistor gives $\dfrac{(59.4\ \text{V})^2}{46\ \Omega} = 76.7$ W. The same heating. That is exactly what the rms value means.

Practise with new numbers

Try it in Python

The program samples a sine wave of peak $325$ V over one full cycle. It finds the plain average, the average of the squares and the rms value.

Common mistakes

• Using the peak value in $P = V^2/R$. That gives the greatest power, which is twice the average power. Use the rms value for the average.
• Thinking the average over a cycle shows the size of an AC voltage. It is zero. The rms value shows the size.
• Writing $V_{rms} = V_m/2$. The right factor is $1/\sqrt{2} = 0.707$.
• Forgetting that a meter reads rms. A reading of $230$ V on the mains is not the peak value.
• Setting the calculator to degrees. In $\sin \omega t$ the angle $\omega t$ is in radians.
• Saying a bulb flickers $50$ times a second at $50$ Hz. It brightens at every positive and every negative peak, so $100$ times a second.

Remember these

• $v = V_m \sin \omega t$, with $\omega = 2\pi f$ and $T = 1/f$.
• Across a resistor, $i = I_m \sin \omega t$ with $I_m = V_m/R$. The current is in phase with the voltage.
• $I_{rms} = \dfrac{I_m}{\sqrt{2}}$ and $V_{rms} = \dfrac{V_m}{\sqrt{2}}$. Meters and ratings give rms values.
• Average power in a resistor: $P = V_{rms} I_{rms} = I_{rms}^2 R = \dfrac{V_{rms}^2}{R}$.
• The mean of a sine wave over a full cycle is zero, and over half a cycle it is $0.637\, I_m$.

Check yourself

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

1. A source gives $v = 40 \sin(200\pi t)$, with $v$ in V and $t$ in s. What are its frequency and its rms value?

2. An AC voltmeter across a lamp reads $24$ V. What is the peak value of the voltage?

3. The average of an alternating current over a cycle is zero. Why does a heater still get hot?

4. The peak current in a lamp of resistance $400\ \Omega$ is $0.50$ A. What is the average power?