RMS and AC Power

Overview

The root mean square value measures the effective heating action of a periodic current or voltage. It is needed because a symmetric alternating current has zero mean over a complete cycle, even though it heats a resistor.

This note develops the idea in the order named by root mean square: square the instantaneous values, find their mean, then take the square root.

Core Ideas

• Squaring removes the sign and matches the $i^{2}R$ dependence of resistive heating.
• RMS is defined for any periodic waveform; the factor $1/\sqrt{2}$ is specific to a sinusoid.
• For sinusoidal current in a pure resistor, mean power is half the maximum instantaneous power.
• Resistive AC power formulae use rms values, not a mixture of peak and rms values.

General rms definition

For a periodic current $i(t)$ of period $T$,

$$I_{\mathrm{rms}}=\sqrt{\frac{1}{T}{\int }_0^T{i}^2(t)\mathrm{d}t}.$$

Equivalently, $I_{\mathrm{rms}}$ is the steady direct current that produces the same mean heating power in the same resistor.

For voltage,

$$V_{\mathrm{rms}}=\sqrt{\frac{1}{T}{\int }_0^T{v}^2(t)\mathrm{d}t}.$$

Figure: The rms operation. First square the current so both half-cycles become positive; then average $i^2$ over one complete period; finally take the square root. For $i=I_0\sin \omega t$, the mean square is $I_0^2/2$, so the rms current is $I_0/\sqrt{2}$.

Sinusoidal rms derivation

Let

$$i=I_0\sin \omega t.$$

Then

$$i^2=I_0^2{\sin }^2\omega t.$$

Over one complete period,

$$\frac{1}{T}{\int }_0^T{\sin }^2\omega t\mathrm{d}t=\frac{1}{2}.$$

Therefore,

$$I_{\mathrm{rms}}=\sqrt{\frac{I_0^2}{2}}=\frac{I_0}{\sqrt{2}}.$$

Similarly, for sinusoidal voltage,

$$V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}.$$

The peak-to-peak value is $2I_0$ or $2V_0$; it is not twice the rms value.

Mean power in a pure resistor

In a pure resistor, voltage and current are in phase. Using

$$i=I_0\sin \omega t,$$

the instantaneous power is

$$p(t)=i^{2}R=I_0^{2}R{\sin }^2\omega t.$$

Figure: The current waveform has period $T$, but squaring produces a non-negative power waveform of period $T/2$. The dashed mean-power line is at half the maximum instantaneous power.

The maximum instantaneous power is

$$p_{\max }=I_0^{2}R.$$

Since the mean of ${\sin }^2\omega t$ is $1/2$,

$$P_{\mathrm{avg}}=\frac{1}{2}{I}_0^{2}R=\frac{1}{2}{p}_{\max }.$$

Using $I_{\mathrm{rms}}=I_0/\sqrt{2}$,

$$P_{\mathrm{avg}}=I_{\mathrm{rms}}^{2}R.$$

For the same pure resistor,

$$P_{\mathrm{avg}}=V_{\mathrm{rms}}{I}_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}^2}{R}.$$

The simple product $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ is being used here for in-phase voltage and current. It is not a universal real-power formula for arbitrary AC components.

Worked Example 1: sinusoidal supply

A $48\Omega$ heater is connected to a $240\mathrm{V}$ rms supply.

$$P_{\mathrm{avg}}=\frac{V_{\mathrm{rms}}^2}{R}=\frac{240^2}{48}=1200\mathrm{W}.$$

No conversion by $\sqrt{2}$ is needed because the stated voltage is already rms.

Worked Example 2: peak voltage given

A sinusoidal source with $V_0=12.0\mathrm{V}$ is connected across a $6.0\Omega$ resistor.

$$V_{\mathrm{rms}}=\frac{12.0}{\sqrt{2}}=8.49\mathrm{V},$$

so

$$P_{\mathrm{avg}}=\frac{8.49^2}{6.0}=12.0\mathrm{W}.$$

The maximum instantaneous power is $p_{\max }=V_0^2/R=24.0\mathrm{W}$, confirming $P_{\mathrm{avg}}=p_{\max }/2$.

Worked Example 3: a non-sinusoidal waveform

A current is $+3.0\mathrm{A}$ for one quarter of every period and zero for the remaining three quarters. Its rms value is

$$I_{\mathrm{rms}}=\sqrt{\frac{(3.0{)}^2(T/4)+0^2(3T/4)}{T}}=1.5\mathrm{A}.$$

This is not $3.0/\sqrt{2}$, because the waveform is not sinusoidal. More generally, for piecewise-constant intervals,

$$I_{\mathrm{rms}}=\sqrt{\frac{\sum i_k^2\Delta t_k}{\sum \Delta t_k}}.$$

Common Mistakes

Averaging before squaring

For a symmetric sinusoid, $⟨i⟩=0$, but $\sqrt{⟨i^2⟩}\ne 0$. Follow the order: square, mean, root.

Applying $1/\sqrt{2}$ to every waveform

The factor follows from the mean of ${\sin }^2$; use the general definition for another shape.

Writing instantaneous power as $P$

Use $p(t)$ for the changing instantaneous power and $P_{\mathrm{avg}}$ for its cycle mean. This prevents confusion between $p_{\max }$ and $P_{\mathrm{avg}}$.

Mixing peak and rms quantities

Do not use $V_0{I}_{\mathrm{rms}}$. Convert first or derive directly from the instantaneous functions.

Exam Relevance

Questions commonly ask you to derive $P_{\mathrm{avg}}=p_{\max }/2$, convert peak and rms values, or determine rms from a graph. State the waveform and load assumptions before selecting a formula.

Links

• Alternating Current
• Alternating Current Common Exam Traps
• DC Circuits

Summary

For any periodic current,

$$I_{\mathrm{rms}}=\sqrt{\frac{1}{T}{\int }_0^T{i}^2\mathrm{d}t}.$$

For a sinusoid only,

$$I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}},V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}.$$

For sinusoidal current in a pure resistor,

$$P_{\mathrm{avg}}=\frac{1}{2}{p}_{\max }=I_{\mathrm{rms}}^{2}R.$$