Alternating currents and rms values

What you'll learn

What an alternating current is and how to read AC graphs.
How peak, period, frequency, and instantaneous values are connected.
Why we use root mean square values for AC.
How to calculate power in AC circuits using rms voltage and rms current.

Starting point: direct current and power

Before alternating current, you already know direct current.

A direct current is a current that flows in one direction only. In a simple DC circuit, the current may be steady, so the current-time graph is a horizontal line.

For a resistor, the usual equations are:

V = I R

I^2R = \frac{V^2}{R}

where:

$V$ is potential difference in volts, V
$I$ is current in amperes, A
$R$ is resistance in ohms, $Ω\Omega$
$P$ is power in watts, W

These equations still matter for AC circuits, but you must be careful about which value of voltage or current you use.

Key Idea

Power depends on current squared

In a resistor, $P = I^2R$ . This matters because even when AC current changes direction, the heating effect does not “cancel out”: squaring the current makes the power positive.

What is an alternating current?

An alternating current, usually shortened to AC, is a current that repeatedly changes direction.

In many A-Level questions, the AC is assumed to be sinusoidal, meaning its graph has the shape of a sine wave.

Definition

Alternating current

An alternating current is an electric current that changes direction periodically.

A sinusoidal alternating current can be written as:

I_0 \sin(2\pi ft)

where:

$I$ is the instantaneous current, the current at one particular time
$I_0$ is the peak current, the maximum value of the current
$f$ is frequency in hertz, Hz
$t$ is time in seconds, s

The same idea applies to voltage:

V_0 \sin(2\pi ft)

where $V_0$ is the peak voltage.

Sinusoidal alternating current graph showing peak current, rms levels, period, and direction changes

Period and frequency

The period, $T$ , is the time taken for one complete cycle of the waveform.

The frequency, $f$ , is the number of complete cycles per second.

They are linked by:

\frac{1}{T}

and therefore:

\frac{1}{f}

For UK mains electricity, the frequency is 50 Hz, meaning there are 50 complete cycles every second.

Example

Finding the period of mains electricity

UK mains electricity has frequency 50 Hz. Find its period.

Use the relationship between frequency and period:
$\frac{1}{f}$
Substitute $50\ \text{Hz}$ :
$\frac{1}{50}$
Calculate the period:
$0.020\ \text{s}$
So one complete cycle takes 0.020 s, or 20 ms.

Common Mistake

Confusing period and half-period

For AC, the current changes direction every half-cycle, but the period is the time for a full cycle. If the current goes from positive peak to negative peak, that is only half a period.

Peak, peak-to-peak, and instantaneous values

The peak value is the maximum magnitude of the voltage or current.

For current, this is usually written $I_0$ . For voltage, it is usually written $V_0$ .

The peak-to-peak value is the difference between the maximum positive value and the maximum negative value.

For a symmetrical sine wave:

Ipeak-to-peak=2I0I_{\text{peak-to-peak}} = 2I_0

and:

Vpeak-to-peak=2V0V_{\text{peak-to-peak}} = 2V_0

The instantaneous value is the value at a particular instant in time. For example, if:

I_0 \sin(2\pi ft)

then $I$ keeps changing as time changes.

Example

Finding an instantaneous current

A sinusoidal AC has peak current 4.0 A and frequency 50 Hz. Find the current at time 5.0 ms after passing through zero in the positive direction.

Write the current equation:
$I_0 \sin(2\pi ft)$
Convert the time into seconds:
$s5.0\ \text{ms} = 5.0 \times 10^{-3}\ \text{s}$
Substitute the values:
$\sin(2\pi \times 50 \times 5.0 \times 10^{-3})$
Evaluate the angle:
$2π×50×5.0×10−3=π22\pi \times 50 \times 5.0 \times 10^{-3} = \frac{\pi}{2}$
so:
$\sin\left(\frac{\pi}{2}\right) = 4.0\ \text{A}$

Tip

Calculator mode

For AC calculations using $2πft2\pi ft$ , your calculator should be in radians if you are using $π\pi$ directly in the sine function.

Why average current is not enough

Over a complete cycle of a sinusoidal AC, the average current is zero. The positive half-cycle and negative half-cycle cancel.

But a lamp connected to AC still lights up. A resistor still heats up.

So average current is not a useful measure of the heating effect of AC.

The useful quantity is the root mean square, or rms, value.

Definition

Root mean square value

The rms value of an alternating current is the value of the steady direct current that would produce the same mean power in a resistor.

That definition is important: rms values are about equivalent heating effect.

RMS current and voltage for a sinusoidal AC

For a sinusoidal alternating current:

Irms=I02I_{\text{rms}} = \frac{I_0}{\sqrt{2}}

For a sinusoidal alternating voltage:

Vrms=V02V_{\text{rms}} = \frac{V_0}{\sqrt{2}}

Rearranging:

I0=2IrmsI_0 = \sqrt{2}I_{\text{rms}}

and:

V0=2VrmsV_0 = \sqrt{2}V_{\text{rms}}

These relationships only apply to a sinusoidal waveform.

Key Idea

RMS is the useful AC value

When an AC voltage or current is quoted without saying “peak”, it is usually an rms value. This is especially true for mains electricity.

Example

Converting rms voltage to peak voltage

UK mains voltage is approximately 230 V rms. Find the peak voltage.

Choose the rms-to-peak relationship for a sinusoidal voltage:
$V0=2VrmsV_0 = \sqrt{2}V_{\text{rms}}$
Substitute $VV_{\text{rms}} = 230\ \text{V}$ :
$V0=2×230V_0 = \sqrt{2} \times 230$
Calculate:
$VV_0 \approx 325\ \text{V}$
So a 230 V rms supply has a peak voltage of about 325 V.

Common Mistake

Thinking 230 V mains means 230 V peak

The quoted UK mains voltage, 230 V, is an rms value. The peak voltage is larger, about 325 V for a sinusoidal supply.

Where the rms formula comes from

For a resistor:

P = I^2R

If the current is sinusoidal:

I_0 \sin(2\pi ft)

so the instantaneous power is:

I_0^2R\sin^2(2\pi ft)

The average value of $sin^2$ over a complete cycle is $12\frac{1}{2}$ , so the mean power is:

Pmean=I02R2P_{\text{mean}} = \frac{I_0^2R}{2}

For a DC current producing the same heating effect:

I_{\text{rms}}^2R

Equating the powers gives:

Irms2R=I02R2I_{\text{rms}}^2R = \frac{I_0^2R}{2}

so:

Irms=I02I_{\text{rms}} = \frac{I_0}{\sqrt{2}}

Voltage and current in phase for a resistor, with instantaneous power always positive and average power shown

Common Mistake

RMS depends on waveform shape

The result $Irms=I02I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ is for a sinusoidal waveform. A square wave, triangular wave, or irregular waveform has a different rms relationship.

Power calculations using rms values

For AC power calculations in a resistor, use rms values in the familiar power equations:

Pmean=VrmsIrmsP_{\text{mean}} = V_{\text{rms}}I_{\text{rms}}

Pmean=Irms2RP_{\text{mean}} = I_{\text{rms}}^2R

Pmean=Vrms2RP_{\text{mean}} = \frac{V_{\text{rms}}^2}{R}

These give the mean power, not the instantaneous power.

Example

Calculating mean power in a resistor

A 12 V rms AC supply is connected across a 6.0 $Ω\Omega$ resistor. Calculate the mean power dissipated.

Choose the power equation using voltage and resistance:
$Pmean=Vrms2RP_{\text{mean}} = \frac{V_{\text{rms}}^2}{R}$
Substitute the values:
$Pmean=1226.0P_{\text{mean}} = \frac{12^2}{6.0}$
Calculate:
$WP_{\text{mean}} = 24\ \text{W}$
The resistor dissipates a mean power of 24 W.

Common Mistake

Using peak values in DC-style power equations

If you put peak voltage into $\frac{V^2}{R}$ , you calculate the peak instantaneous power, not the mean power. For average heating power, use rms values.

RMS current from peak current

Sometimes the question gives a peak current and asks for power.

The safest route is usually:

Convert peak current to rms current.
Use $Pmean=Irms2RP_{\text{mean}} = I_{\text{rms}}^2R$ .

Example

Mean power from peak current

A sinusoidal alternating current has peak current 3.0 A through a 10 $Ω\Omega$ resistor. Calculate the mean power dissipated.

Convert peak current to rms current:
$Irms=I02=3.02I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{3.0}{\sqrt{2}}$ $AI_{\text{rms}} \approx 2.12\ \text{A}$
Use the mean power equation for a resistor:
$Pmean=Irms2RP_{\text{mean}} = I_{\text{rms}}^2R$
Substitute the values:
$Pmean=2.122×10P_{\text{mean}} = 2.12^2 \times 10$
Calculate:
$WP_{\text{mean}} \approx 45\ \text{W}$

Tip

A quicker route

For a sinusoidal current in a resistor, you can also use $Pmean=I02R2P_{\text{mean}} = \frac{I_0^2R}{2}$ . This works because $Irms2=I022I_{\text{rms}}^2 = \frac{I_0^2}{2}$ .

Reading AC graphs

On an AC graph, check carefully what the vertical axis shows.

It could show:

instantaneous voltage, $V$
instantaneous current, $I$
peak value, $V_0$ or $I_0$
rms value, $VrmsV_{\text{rms}}$ or $IrmsI_{\text{rms}}$

The graph itself normally shows instantaneous values. RMS values are not usually points on the sine curve; they are equivalent DC values based on power.

Example

Using a graph to find rms current

An AC current graph has a maximum value of 8.0 A. The waveform is sinusoidal. Find the rms current.

Identify the maximum value as the peak current:
$AI_0 = 8.0\ \text{A}$
Use the rms relationship for a sinusoidal current:
$Irms=I02I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$
Substitute and calculate:
$AI_{\text{rms}} = \frac{8.0}{\sqrt{2}} \approx 5.7\ \text{A}$

Exam technique

In the exam

Decide whether the value given is peak, peak-to-peak, or rms before substituting into equations.
For mean power in a resistor, use rms values: $Pmean=VrmsIrms=Irms2RP_{\text{mean}} = V_{\text{rms}}I_{\text{rms}} = I_{\text{rms}}^2R$ .
Remember that $Vrms=V02V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$ and $Irms=I02I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ only apply to sinusoidal AC.
If you see frequency, period, or time on a graph, use $\frac{1}{T}$ and check whether the interval shown is a full cycle or half-cycle.

Self review

Check yourself

Why is the average current over one complete AC cycle zero, but the average power is not zero?
A sinusoidal supply has peak voltage 20 V. What equation would you use to find its rms voltage?
Why should rms values be used when calculating the mean power dissipated in a resistor?

Alternating currents and rms values

What you'll learn

Starting point: direct current and power

What is an alternating current?

Period and frequency

Peak, peak-to-peak, and instantaneous values

Why average current is not enough

RMS current and voltage for a sinusoidal AC

Where the rms formula comes from

Power calculations using rms values

RMS current from peak current

Reading AC graphs

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Alternating currents and rms values Revision Guide