# Power in ac circuits.

## Occasional notes in electronics series.

Are you perplexed and confused about using complex phasor analysis to represent real quantities such as Voltage, Current, and Power? If so, read on....

## Complex current and voltage.

Complex numbers are used in ac circuits to represent both the amplitude and the phase of an ac voltage or current.

The principal advantages to the electronics engineer lie in writing down the equations with a great deal more compactness; in making many numerical shortcuts in handling voltage current and power in circuits having both resistance and reactance; and in gaining insight into creative circuit methods of handling alternating signals.

In general the time-development of an ac sinusoidal signal (voltage or current) may be written as

```
A cos(omega t + phi)

or

A sin(omega t + phi)

where omega = (2 pi f) is the angular frequency in radians per second
and (phi) is the phase angle in radians. A is of course the size or
peak amplitude. Also, f is the frequency in Hz or cycles per second.
It also tells us the number of rotations of the complex phasor, to
be defined below, around the origin of the Argand diagram,
in one second. Omega tells us the number of radians of angle
the complex phasor sweeps out in one second.

```
Here, everything written down is a real quantity. If we allow the use of complex numbers, which are really just a way of writing down two real numbers with one symbol, we can absorb the phase information phi and theta into a "complex phasor" or "complex amplitude" a.

We are going to define the complex amplitude a as

```
a = A exp(j phi)

and by the usual relations for complex numbers, we can write this as

a = A cos(phi) + jA sin(phi)

If by the same token we represent the time development of the ac
signal as

exp(j omega t)

then multiplying by the complex amplitude a we find that the
signal is represented as

A exp(j phi) exp (j omega t)

which is just

A exp [j(omega t + phi)]

and if we take the real part of this we recover the original
real relationship for the signal

A cos(omega t + phi).

The imaginary part is of course

A sin(omega t + phi)

```
Thus a shorthand way of writing our original signal A cos(omega t + phi) is as "a exp(j omega t)" and here we assume that to recover the actual signal, the phase angle phi is the phase angle of the complex phasor or amplitude a, and it is assumed that we take the Real part of the expression "a exp(j omega t)" to recover the entirely real voltage or current.

Thus, given the complex amplitude a, the size of the ac signal is given by the complex modulus of a, which is written mod(a) or just |a|, and this is just our original real quantity A.

On the other hand the phase angle phi is given by the phase angle of a, called the "argument" of a and written arg(a).

```

Since a = A cos(phi) + jA sin(phi)

we see that the real part of a is A cos(phi), written as Re(a),
and the imaginary part of a is A sin(phi), written as Im(a).

Clearly, Im(a)/Re(a) = sin(phi)/cos(phi) = tan(phi)

and so   phi = arc tan (Im(a)/Re(a))

Also, A^2 = [Re(a)]^2 + [Im(a)]^2 from Pythagoras' theorem
and using the relationship sin^2 + cos^2 = 1.

```
From these relations we can get the real and imaginary parts of the complex amplitude a related to the size of the signal and its phase angle. This is called a "rectangular to polar co-ordinate transformation" and is best seen graphically on the Argand diagram. Often it is possible to simplify calculations by doing repeated transformations between the polar and rectangular forms of the complex quantities, in both directions as needed.

## Power in complex notation

Suppose we have a current of size I (peak ac value) flowing through a resistor R in series with a reactance X. The average power dissipated is just I^2 R/2, because the same current flows through both the resistance and the reactance, and the reactance dissipates no power. The factor 2 arises because the average value of the square of a unit sinusoidal signal over one cycle is just 1/2. In determining the average power dissipated we intrinsically assume the average is taken over a whole number of cycles of the ac signal.

Now suppose the complex current phasor is i = I exp(j phi). Taking the complex conjugate (denoted by i* and got by replacing j by -j) we find i* = I exp(-j phi) and that I^2 = ii*.

Thus the average power dissipated is numerically equal to (1/2)ii*R. This is a real quantity.

Now if we were to write the impedance R+jX = Z, and noting that ii* is real, then the power dissipated is numerically equal to the real part of Zii*/2. Replacing the quantity Zi by the complex voltage phasor v, we see transparently that the power dissipated is Re[vi*/2]. This is a very useful relationship as we don't have to know the impedance at all.

Tutorial exercise for the reader. Prove that Re[vi*/2] = Re[v*i/2]. This is another useful result. Note that if v = Zi then v* = Z*i*.

As a method of remembering the result, recite a number of times "The average power delivered to a port in an ac circuit is half the real part of vee eyestar".

Of course, the quantities i and v represent the peak values of the oscillating current and voltage respectively, which is why it is necessary to have the factor of 1/2. This factor is missing if the current and voltage amplitudes are r.m.s. values.

An example. A capacitor of 1 microFarad is connected in series with a resistance of 1 ohm to a mains supply having r.m.s. voltage 240 volts at a frequency of 50 Hz. What is the power drawn from the mains and where does this power appear?

• Solution. The reactance of the 1 microfarad capacitor at 50 Hz is just 1/(2.pi.50.1E-6) = 3183 ohms. The impedance of the load across the mains is therefore Z = 1 + 3183j ohms and the current I is therefore 240/(1+3183j) amps = 2.37E-5 - 0.0754j amps. Since these are r.m.s.quantities we don't need the factor of 1/2, so the power taken from the supply is Re(VI*) = 240 times 2.357E-5 or 5.7 milliwatts. This is also the result we get if we approximate the impedance across the supply by 3183+1 ohms = 3184 ohms, and find the r.m.s. current from 240/3184 = 75.7 milliamps. Then the power estimated from (75.7).(75.7).1E-6 which is just R(I^2) works out the same at 5.7 milliwatts. The discrepancy due to the approximation is in the next significant figure of the result. The power of course appears in the 1 ohm resistor.

Your turn. Work out the power dissipated in a 47 ohm resistor connected through a 25 microfarad capacitor to a 60Hz supply at 117 volts r.m.s.