Power and RMS values

Power and RMS values

The power p converted in a resistor (ie the rate of conversion of electrical energy to heat) is p(t) = iv = v2/R = i2R. We use lower case p(t) because this is the expression for the instantaneous power at time t. Usually, we are interested in the mean power delivered, which is normally written P. P is the total energy converted in one cycle, divided by the period T of the cycle, so:

In the last line, we have used a standard trigonometrical identity that   cos(2A) = 1 - 2 sin²A. Now the sinusoidal term averages to zero over any number of complete cycles, so the integral is simple and we obtain

This last set of equations are useful because they are exactly those normally used for a resistor in DC electricity. However, one must remember that P is the average power, and V = V_m/2^1/2 and I= I_m/2^1/2. Looking at the integral above, and dividing by R, we see that I is equal to the square root of the mean value of i², so I is called the root-mean-square or RMS value. Similarly, V = V_m/2^1/2 ~ 0.71*V_m is the RMS value of the voltage.
When talking of AC, RMS values are so commonly used that, unless otherwise stated, you may assume that RMS values are intended*. For instance, normal domestic AC in Australia is 240 Volts AC with frequency 50 Hz. The RMS voltage is 240 volts, so the peak value V_m= V.2^1/2 = 340 volts. So the active wire goes from +340 volts to -340 volts and back again 50 times per second. (This is the answer to the teaser question at the top of the page: rectification of the 240 V mains can give both + 340 Vdc and -340 Vdc.)
* An exception: manufacturers and sellers of HiFi equipment sometimes use peak values rather than RMS values, which makes the equipment seem more powerful than it is.
Power in a resistor. In a resistor R, the peak power (achieved instantaneously 100 times per second for 50 Hz AC) is V_m²/R = i_m²*R. As discussed above, the voltage, current and so the power pass through zero volts 100 times per second, so the average power is less than this. The average is exactly as shown above: P = V_m²/2R = V²/R.
Power in inductors and capacitors. In ideal inductors and capacitors, a sinusoidal current produces voltages that are respecively 90° ahead and behind the phase of the current. So if i = I_msin wt, the voltages across the inductor and capacitor are V_mcos wt and -V_mcos wt respectively. Now the integral of cos*sin over a whole number of cycles is zero. Consequently, ideal inductors and capacitors do not, on average, take power from the circuit.