RMS and power in single and three phase AC circuits
Power in AC circuits, the use of RMS quantities and 3 phase AC – including answers to these questions:
 What are RMS values?
 How can you work out the power developed in an AC circuit?
 How can you get 680 V dc from a 240 V ac supply just by rectifying?
 When do you need three phases and why do you need four wires?


This page provides answers to these questions. This is a resource page from
Physclips. It is a subsidiary page to
the main
AC circuits site.
There are separate pages on
RC filters, integrators
and differentiators,
LC oscillations and
motors
and generators.
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 = v^{2}/R = i^{2}R.
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 sin^{2}A. 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 and I= I_{m}/√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 i^{2}, so I is called the
rootmeansquare or RMS value. Similarly, V = V_{m}/√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 = 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_{m}^{2}/R = i_{m}^{2}*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_{m}^{2}/2R = V^{2}/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_{m}sin wt, the voltages across the inductor and capacitor are V_{m}cos wt
and V_{m}cos 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.
Three phase AC
Single phase AC has the advantage that it only requires
2 wires. Its disadvantage is seen in the graph at the top of this page:
twice every cycle V goes to zero. If you connect a phototransistor
circuit to an oscilloscope, you will see that fluorescent lights turn
off 100 times per second (or 120, if you are on 60 Hz supply).
What if you need a more even supply of electricity? One can store energy
in capacitors, of course, but with high power circuits this would require
big, expensive capacitors. What to do?
An AC
generator may have more than one coil. If it has three
coils, mounted at relative angles of 120°, then it will produce
three sinusoidal emfs with relative phases of 120°, as shown
in the upper figure at right. The power delivered to a resistive
load by each of these is proportional to V^{2}. The
sum of the three V^{2} terms is a constant. We saw
above that the average of V^{2} is half the peak
value, so this constant is 1.5 times the peak amplitude for
any one circuit, as is shown in the lower figure at right.
Do you need four wires? In principle, no. The sum of the three
V terms is zero so, provided that the loads on each phase are
identical, the currents drawn from the three lines add to zero.
In practice, the current in the neutral wire is usually not
quite zero. Further, it should be the same guage as the other
wires because, if one of the loads were to fail and form an
open circuit, the neutral would carry a current similar to
that in the remaining two loads.

The voltage (top) and square of the voltage (bottom)
in the three active lines of 3 phase supply.

Go to the main AC circuits site,
RC filters, integrators
and differentiators
LC oscillations,
or to
Motors and generators.