Mean power in AC circuits uses RMS values

Alternating Currents — AQA A-Level Physics

Since RMS values are the DC equivalents, you can use them directly in the standard powerThe rate of energy transfer. Measured in watts (W). equations:

P_{\text{mean}} = I_{\text{rms}} \times V_{\text{rms}}

$P_{\text{mean}}$: mean (average) powerThe rate of energy transfer. Measured in watts (W). dissipated (W)
$I_{\text{rms}}$: RMS currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). (A)
$V_{\text{rms}}$: RMS voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference. (V)

You can also write this in terms of peak values. Since $I_{\text{rms}} = I_0 / \sqrt{2}$ and $V_{\text{rms}} = V_0 / \sqrt{2}$:

P_{\text{mean}} = \frac{I_0 V_0}{2}

$P_{\text{mean}}$: mean powerThe rate of energy transfer. Measured in watts (W). (W)
$I_0$: peak currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). (A)
$V_0$: peak voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. (V)

All the usual combinations still work with RMS values: $P = I_{\text{rms}}^2 R$ and $P = V_{\text{rms}}^2 / R$.
The peak (instantaneous maximum) power is $P_{\text{peak}} = I_0 V_0$. The mean power is exactly half of this for a sinusoidal AC signal.
This factor of 2 comes directly from the 1/√2 in each RMS conversion: $(1/\sqrt 2) \times (1/\sqrt 2) = 1/2.$

Worked Example

A heater has a resistance of 48 Ω and is connected to the 230 V mains supply. Calculate (a) the RMS current and (b) the mean power dissipated.

Show Solution

List known values

RMS voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference.: $V_{\text{rms}} = 230 \text{ V}$ (mains is quoted as RMS)
ResistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω).: $R = 48 \text{ } \Omega$

Calculate the RMS current

Using Ohm's lawFor an ohmic conductor at constant temperature, current is directly proportional to potential difference. with RMS values:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{230}{48}$$ $$= 4.79 \text{ A}$$

Calculate the mean power

$$P_{\text{mean}} = I_{\text{rms}} \times V_{\text{rms}} = 4.79 \times 230$$ $$= 1100 \text{ W}$$

Alternatively: $P = V_{\text{rms}}^2 / R = 230^2 / 48 = 1102 \text{ W}$ (same answer, minor rounding difference).

Answer

(a) $I_{\text{rms}} = 4.8$ A (2 s.f.) (b) $P_{\text{mean}} = 1100$ W (2 s.f.)

Common Mistake MEDIUM

Students often: Using peak values in power equations and getting double the actual power.
Instead: $P = IV$ only gives mean power if you use RMS values. If you use peak values, you get peak power, which is twice the mean. For mean power from peaks: $P_{\text{mean}} = I_{0} V_{0}/2$.