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.

RC Series combinations

RC Series combinations

When we connect components together, Kirchoff's laws apply at any instant. So the voltage v(t) across a resistor and capacitor in series is just

v_series(t) = v_R(t) + v_C(t)

however the addition is complicated because the two are not in phase. The next animation makes this clear: they add to give a new sinusoidal voltage, but the amplitude is less than V_mR(t) + V_mC(t). Similarly, the AC voltages (amplitude times √2) do not add up. This may seem confusing, so it's worth repeating:

v_series = v_R + v_C        but V_series < V_R + V_C.

This should be clear on the animation and the still graphic below: check that the voltages v(t) do add up, and then look at the magnitudes. The amplitudes and the RMS voltages V do not add up in a simple arithmetical way.

Here's where phasor diagrams are going to save us a lot of work. Play the animation again (click play), and look at the projections on the vertical axis. Because we have sinusoidal variation in time, the vertical component (magnitude times the sine of the angle it makes with the x axis) gives us v(t). But the y components of different vectors, and therefore phasors, add up simply: if

r_total = r₁ + r₂,        then r_{y total} = r_y1 + r_y2.

So v(t), the sum of the y projections of the component phasors, is just the y projection of the sum of the component phasors. So we can represent the three sinusoidal voltages by their phasors. (While you're looking at it, check the phases. You'll see that the series voltage is behind the current in phase, but the relative phase is somewhere between 0 and 90°, the exact value depending on the size of V_R and V_C. We'll discuss phase below.)Now let's stop that animation and label the values, which we do in the still figure below. All of the variables (i, v_R, v_C, v_series) have the same frequency f and the same angular frequency ω, so their phasors rotate together, with the same relative phases. So we can 'freeze' it in time at any instant to do the analysis. The convention I use is that the x axis is the reference direction, and the reference is whatever is common in the circuit. In this series circuit, the current is common. (In a parallel circuit, the voltage is common, so I would make the voltage the horizontal axis.) Be careful to distinguish v and V in this figure!

(Careful readers will note that I'm taking a shortcut in these diagrams: the size of the arrows on the phasor diagrams are drawn the same as the amplitudes on the v(t) graphs. However I am just calling them V_R, V_C etc, rather than V_mR, V_mR etc. The reason is that the peak values (V_mR etc) are rarely used in talking about AC: we use the RMS values, which are peak values times 0.71. Phasor diagrams in RMS have the same shape as those drawn using amplitudes, but everything is scaled by a factor of 0.71 = 1/√2.)The phasor diagram at right shows us a simple way to calculate the series voltage. The components are in series, so the current is the same in both. The voltage phasors (brown for resistor, blue for capacitor in the convention we've been using) add according to vector or phasor addition, to give the series voltage (the red arrow). By now you don't need to look at v(t), you can go straight from the circuit diagram to the phasor diagram, like this:

From Pythagoras' theorem:

V²_mRC = V²_mR + V²_mC

If we divide this equation by two, and remembering that the RMS value V = V_m/√2, we also get:

Now this looks like Ohm's law again: V is proportional to I. Their ratio is the series impedance, Z_series and so for this series circuit,

    

Note the frequency dependence of the series impedance Z_RC: at low frequencies, the impedance is very large, because the capacitive reactance 1/ωC is large (the capacitor is open circuit for DC). At high frequencies, the capacitive reactance goes to zero (the capacitor doesn't have time to charge up) so the series impedance goes to R. At the angular frequency ω = ω_o = 1/RC, the capacitive reactance 1/ωC equals the resistance R. We shall show this characteristic frequency* on all graphs on this page.

* Let's check the units: it should be quick. First, a radian is the ratio of an arc to a radius, so it is a pure number. R is in ohms, and we can use Ohm's law (V = IR) and the definition of current (I = dQ/dt) to write ohm = volt/ampere = volt/(coulomb/second) = volt*second/coulomb. C (charge per unit voltage) is in farads (coulomb/volt), so RC has the units of (volt*second/coulomb)(coulomb/volt) = second. See, it was quick: it only took us 30 ohm.farads!

Remember how, for two resistors in series, you could just add the resistances: R_series = R₁ + R₂ to get the resistance of the series combination. That simple result comes about because the two voltages are both in phase with the current, so their phasors are parallel. Because the phasors for reactances are 90° out of phase with the current, the series impedance of a resistor R and a reactance X are given by Pythagoras' law:

Z_series² = R² + X² .

Ohm's law in AC. We can rearrange the equations above to obtain the current flowing in this circuit. Alternatively we can simply use the Ohm's Law analogy and say that I = V_source/Z_RC. Either way we get

     

where the current goes to zero at DC (capacitor is open circuit) and to V/R at high frequencies (no time to charge the capacitor).So far we have concentrated on the magnitude of the voltage and current. We now derive expressions for their relative phase, so let's look at the phasor diagram again.

From simple trigonometry, the angle by which the current leads the voltage is

    tan^-1 (V_C/V_R) = tan^-1 (IX_C/IR)  = tan^-1 (1/ωRC) = tan^-1 (1/2πfRC).

However, we shall refer to the angle φ by which the voltage leads the current. The voltage is behind the current because the capacitor takes time to charge up, so φ is negative, ie

φ = -tan^-1 (1/ωRC) = tan^-1 (1/2πfRC).    

At low frequencies, the impedance of the series RC circuit is dominated by the capacitor, so the voltage is 90° behind the current. At high frequencies, the impedance approaches R and the phase difference approaches zero. The frequency dependence of Z and φ are important in the applications of RC circuits. The voltage is mainly across the capacitor at low frequencies, and mainly across the resistor at high frequencies. Of course the two voltages must add up to give the voltage of the source, but they add up as vectors.

V²_RC = V²_R + V²_C.

At the frequency ω = ω_o = 1/RC, the phase φ = 45° and the voltage fractions are V_R/V_RC = V_C/V_RC = 1/2V^1/2 = 0.71.

So, by chosing to look at the voltage across the resistor, you select mainly the high frequencies, across the capacitor, you select low frequencies. This brings us to one of the very important applications of RC circuits, and one which merits its own page:  where we use sound files as examples of RC filtering.

Three phase AC induction motors

Three phase AC induction motors

Single phase is used in domestic applications for low power applications but it has some drawbacks. One is that it turns off 100 times per second (you don't notice that the fluorescent lights flicker at this speed because your eyes are too slow: even 25 pictures per second on the TV is fast enough to give the illusion of continuous motion.) The second is that it makes it awkward to produce rotating magnetic fields. For this reason, some high power (several kW) domestic devices may require three phase installation. Industrial applications use three phase extensively, and the three phase induction motor is a standard workhorse for high power applications. The three wires (not counting earth) carry three possible potential differences which are out of phase with each other by 120°, as shown in the animation below. Thus three stators give a smoothly rotating field.

If one puts a permanent magnet in such a set of stators, it becomes a synchronous three phase motor. The animation shows a squirrel cage, in which for simplicity only one of the many induced current loops is shown. With no mechanical load, it is turning virtually in phase with the rotating field. The rotor need not be a squirrel cage: in fact any conductor that will carry eddy currents will rotate, tending to follow the rotating field. This arrangement can give an induction motor capable of high efficiency, high power and high torques over a range of rotation rates.