Peak values, rms values and power in AC

Key ideas

Alternating current swings sinusoidally:

$I = I_{MAX}\sin\omega t \qquad V = V_{MAX}\sin\omega t \qquad \omega = 2\pi f$

$I_{MAX}, V_{MAX}$ — the peak values
$\omega$ — the angular frequency (rad s⁻¹); NZ mains is $f = 50$ Hz
The average of a sine wave is zero, so averages are useless for power. Instead we use the rms (root-mean-square) value — the steady DC value that would heat a resistor at the same average rate:

$V_{MAX} = \sqrt{2}\,V_{rms} \qquad I_{MAX} = \sqrt{2}\,I_{rms}$

NZ mains "230 V" is the rms value; the peak is $230\sqrt{2} \approx 325$ V.
Average power in a resistor uses rms values directly: $P = V_{rms}I_{rms} = I_{rms}^2R$ .
An AC supply of $V_{rms} = 230$ V delivers exactly the same average heating as a 230 V battery — that equivalence is the definition of rms.

Worked ExamplePeak values of NZ mains

A heater draws $6.5$ A rms from the $230$ V rms NZ mains. Find the peak voltage, the peak current, and the average power.

Step 1 — Peak values are √2 times rms

$V_{MAX} = \sqrt{2}(230) = 325 \text{ V} \qquad I_{MAX} = \sqrt{2}(6.5) = 9.2 \text{ A}$

Step 2 — Average power from rms values

$P = V_{rms}I_{rms} = (230)(6.5) = 1.5\times10^{3} \text{ W}$

Tips

Meter readings and appliance ratings are rms unless stated otherwise — that's the convention the whole industry uses.
Insulation and component ratings must survive the peak (325 V for NZ mains), which is why capacitors rated "250 V" fail on mains — always check which value a question wants.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

An AC supply has an rms voltage of $12$ V.

Show that its peak voltage is about $17$ V.

Merit

An AC current is $I = 4.2\sin(100\pi t)$ A.

Determine the rms current and the frequency.

Excellence

Explain what the rms value of an alternating current means physically, and why it — rather than the peak or the average — is used to state power ratings.

Peak values, rms values and power in AC

Covers AS91526

Key ideas

Alternating current swings sinusoidally:

$I = I_{MAX}\sin\omega t \qquad V = V_{MAX}\sin\omega t \qquad \omega = 2\pi f$

$I_{MAX}, V_{MAX}$ — the peak values
$\omega$ — the angular frequency (rad s⁻¹); NZ mains is $f = 50$ Hz
The average of a sine wave is zero, so averages are useless for power. Instead we use the rms (root-mean-square) value — the steady DC value that would heat a resistor at the same average rate:

$V_{MAX} = \sqrt{2}\,V_{rms} \qquad I_{MAX} = \sqrt{2}\,I_{rms}$

NZ mains "230 V" is the rms value; the peak is $230\sqrt{2} \approx 325$ V.
Average power in a resistor uses rms values directly: $P = V_{rms}I_{rms} = I_{rms}^2R$ .
An AC supply of $V_{rms} = 230$ V delivers exactly the same average heating as a 230 V battery — that equivalence is the definition of rms.

Worked ExamplePeak values of NZ mains

A heater draws $6.5$ A rms from the $230$ V rms NZ mains. Find the peak voltage, the peak current, and the average power.

Step 1 — Peak values are √2 times rms

$V_{MAX} = \sqrt{2}(230) = 325 \text{ V} \qquad I_{MAX} = \sqrt{2}(6.5) = 9.2 \text{ A}$

Step 2 — Average power from rms values

$P = V_{rms}I_{rms} = (230)(6.5) = 1.5\times10^{3} \text{ W}$

Tips

Meter readings and appliance ratings are rms unless stated otherwise — that's the convention the whole industry uses.
Insulation and component ratings must survive the peak (325 V for NZ mains), which is why capacitors rated "250 V" fail on mains — always check which value a question wants.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

An AC supply has an rms voltage of $12$ V.

Show that its peak voltage is about $17$ V.

Merit

An AC current is $I = 4.2\sin(100\pi t)$ A.

Determine the rms current and the frequency.

Excellence

Explain what the rms value of an alternating current means physically, and why it — rather than the peak or the average — is used to state power ratings.