Alternating Current

Lessons

Senior 6 · Physics

Alternating Current

4 lessons

Lesson 4 of 4

Power in AC Circuits and Applications

25 min

Why the grid cares about the angle

The Rwanda Grid Code asks distribution companies and large customers to keep their power factor at 0.90 or higher during peak demand. A factory pays for the energy its machines use, so why should REG care whether the factory's current is in step with the voltage? Because a current that is out of phase delivers less energy per ampere, yet every ampere still heats the cables and transformers that carry it. This lesson finds the power in an AC circuit, then uses everything in the unit to explain three technologies: power transmission, AC motors and radio.

The average power in an AC circuit

Let the voltage be $v = V_0\sin\omega t$ and the current $i = I_0\sin(\omega t - \phi)$ , lagging by $\phi$ . The instantaneous power is

p = vi = V_0I_0\sin\omega t\,\sin(\omega t - \phi)

Using $\sin A\sin B = \frac{1}{2}\left[\cos(A - B) - \cos(A + B)\right]$ :

p = \frac{1}{2}V_0I_0\left[\cos\phi - \cos(2\omega t - \phi)\right]

The second term oscillates at twice the supply frequency and averages to zero over a cycle. What remains is

P = \langle p \rangle = \frac{1}{2}V_0I_0\cos\phi = \frac{V_0}{\sqrt{2}}\cdot\frac{I_0}{\sqrt{2}}\cos\phi = V_{\text{rms}}I_{\text{rms}}\cos\phi

Figure 4.1: When the current lags by 60°, the power is negative for part of each cycle; the mean power is $V_{\text{rms}}I_{\text{rms}}\cos 60^\circ$ , half of what the same currents would give in phase.

Check the special cases from Lesson 2: a resistor ( $\phi = 0$ ) gives $P = V_{\text{rms}}I_{\text{rms}}$ ; a pure inductor or capacitor ( $\phi = \pm 90^\circ$ ) gives $P = 0$ . In a series circuit, $\cos\phi = \frac{R}{Z}$ , so $P = I_{\text{rms}}^2Z \cdot \frac{R}{Z} = I_{\text{rms}}^2R$ : only the resistance takes energy.

Definition

The true (active) power $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$ , in watts (W), is the average rate at which energy is transferred. The apparent power $S = V_{\text{rms}}I_{\text{rms}}$ , in volt-amperes (VA), is what the voltage and current would deliver if they were in phase. The reactive power $Q = V_{\text{rms}}I_{\text{rms}}\sin\phi$ , in volt-amperes reactive (var), is energy that surges back and forth each cycle. The power factor is $\cos\phi = \frac{P}{S}$ ; it is called lagging when the current lags (inductive loads) and leading when it leads. Example: 230 V and 4.0 A at $\phi = 30^\circ$ give $S = 920\,\mathrm{VA}$ , $P = 797\,\mathrm{W}$ , power factor 0.87 lagging.

Worked example 1: power in the RLC circuit

In Lesson 3, Worked example 3, a series circuit with $R = 40\,\mathrm{\Omega}$ and $Z = 76.2\,\mathrm{\Omega}$ carried 0.315 A from a 24 V supply. Find the power factor, the apparent power, the true power and the reactive power.

Solution

Power factor $= \cos\phi = \frac{R}{Z} = \frac{40}{76.2} = 0.525$ , leading (the circuit was capacitive).

$S = VI = 24 \times 0.315 = 7.56\,\mathrm{VA}$ ; $P = S\cos\phi = 7.56 \times 0.525 = 3.97\,\mathrm{W}$ .

Check: $P = I^2R = 0.315^2 \times 40 = 3.97\,\mathrm{W}$ . And $Q = S\sin\phi = 7.56 \times \sin 58.4^\circ = 6.44\,\mathrm{var}$ ; note $\sqrt{P^2 + Q^2} = \sqrt{3.97^2 + 6.44^2} = 7.56 = S$ .

Power factor correction

$P$ , $Q$ and $S$ form a right-angled power triangle, $S^{2} = P^{2} + Q^{2}$ . Motors, transformers and fluorescent-lamp ballasts are inductive, so industrial loads usually have a lagging power factor. For the same true power, a low power factor means a larger current, $I = \frac{P}{V\cos\phi}$ , and so larger $I^{2} R$ losses in every cable and transformer between the power station and the load. The supplier must also build equipment rated for the full apparent power.

The cure uses Lesson 2: a capacitor draws a current that LEADS the voltage, which cancels part of the lagging current of the inductive load. Connecting capacitors in parallel with the load reduces $Q$ and $\phi$ without changing $P$ . This is power factor correction, and it is why the Rwanda Grid Code can require 0.90.

Figure 4.2: Power triangles for Worked example 2, before and after correction.

Worked example 2: correcting a workshop's power factor

A carpentry workshop's motors take 5.0 kW at a power factor of 0.70 lagging from a 230 V, 50 Hz supply (illustrative values). (a) Find the current. (b) Find the capacitance that must be connected in parallel to raise the power factor to 0.95. (c) By what fraction are the losses in the supply cable reduced?

Solution

(a) $I_1 = \frac{P}{V\cos\phi_1} = \frac{5000}{230 \times 0.70} = 31.1\,\mathrm{A}$ .

(b) $\phi_1 = \cos^{-1}0.70 = 45.6^\circ$ , so $Q_1 = P\tan\phi_1 = 5000 \times 1.020 = 5.10\,\mathrm{kvar}$ . For 0.95, $\phi_2 = 18.2^\circ$ and $Q_2 = 5000 \times 0.329 = 1.64\,\mathrm{kvar}$ . The capacitor must supply $Q_C = 5.10 - 1.64 = 3.46\,\mathrm{kvar}$ . For a capacitor, $Q_C = \frac{V^2}{X_C} = V^2\omega C$ , so

C = \frac{Q_C}{\omega V^2} = \frac{3458}{2\pi \times 50 \times 230^2} = 2.08 \times 10^{-4}\,\mathrm{F} = 208\,\mathrm{\mu F}

(c) New current $I_2 = \frac{5000}{230 \times 0.95} = 22.9\,\mathrm{A}$ . Losses go as $I^{2}$ : $\left(\frac{22.9}{31.1}\right)^2 = 0.54$ , so the cable losses fall by about 46%, for the same useful power.

Warning

Safety. Power-factor capacitors store charge at mains voltage and can hold a dangerous charge after the supply is switched off. They are fitted with discharge resistors and must be installed and serviced only by a qualified electrician. Never touch the terminals of a large capacitor.

Application 1: transmitting power

Unit 3 showed why the grid transmits at high voltage: for a power $P$ at voltage $V$ , the line loss $I^{2} R$ falls as $\frac{1}{V^2}$ , and only AC can be stepped up and down by transformers. With a power factor, the current is $I = \frac{P}{V\cos\phi}$ and the loss in a line of resistance $R$ becomes

P_{\text{loss}} = \frac{P^2R}{V^2\cos^2\phi}

So the loss depends on the power factor as strongly as on the voltage. (Real grids use three phases; the single-line model here keeps the physics clear.)

Worked example 3: the power factor on a 30 kV feeder

A 30 kV distribution line of total resistance $2.0\,\mathrm{\Omega}$ delivers 10 MW (a simplified single-line model with illustrative values). Find the current and the power lost in the line when the load's power factor is (a) 0.95 and (b) 0.75.

Solution

(a) $I = \frac{10 \times 10^{6}}{30 \times 10^{3} \times 0.95} = 351\,\mathrm{A}$ ; $P_{\text{loss}} = 351^2 \times 2.0 = 2.46 \times 10^{5}\,\mathrm{W} = 246\,\mathrm{kW}$ (2.5% of the power delivered).

(b) $I = \frac{10 \times 10^{6}}{30 \times 10^{3} \times 0.75} = 444\,\mathrm{A}$ ; $P_{\text{loss}} = 444^2 \times 2.0 = 395\,\mathrm{kW}$ (4.0%).

The poorer power factor wastes 149 kW more for the same delivered power: $\left(\frac{0.95}{0.75}\right)^2 = 1.6$ times the loss.

Application 2: AC motors

REG's low-voltage network is designed for 400 V/230 V: three live conductors (phases) carry sinusoidal voltages of 230 V rms relative to neutral, each 120° out of phase with the next. The voltage between two phase conductors is the phasor DIFFERENCE of two phase voltages.

Figure 4.3: The three phase voltages and the line voltage $V_{AB}$ .

The rotating field. Three coils set 120° apart around a motor's stator, fed by the three phases, produce magnetic fields that peak one after another. Their sum is a field of constant strength that ROTATES at the synchronous speed $n_s = \frac{f}{p}$ revolutions per second, where $p$ is the number of pole pairs: 3000 rev/min for one pair on 50 Hz, 1500 rev/min for two pairs.

The induction motor. The rotor is a cylinder of iron laminations with copper or aluminium bars shorted together at the ends; it has no electrical connection to the supply. The rotating field sweeps past the bars and induces currents in them (Unit 3), and by Lenz's law the force on these currents drags the rotor round after the field. The rotor always turns a little slower than the field, otherwise there would be no relative motion and no induced current. The fraction by which it lags is the slip, $s = \frac{n_s - n}{n_s}$ . A four-pole motor on 50 Hz that runs at 1440 rev/min has $n_s = 1500\,\mathrm{rev/min}$ and $s = \frac{60}{1500} = 0.04$ , or 4%.

Single-phase motors in fans and pumps have a second winding in series with a capacitor. The capacitor shifts that winding's current by nearly 90° (Lesson 2), so the two windings together make a rotating field from a single-phase supply. That is the capacitor of Lesson 2, Worked example 2.

Worked example 4: where "400 V" comes from

Show that the voltage between two phases of a 230 V three-phase supply is about 400 V.

Solution

In Figure 4.3, $V_{A}$ and $V_{B}$ each have length 230 V and are 120° apart. The line voltage $V_{AB} = V_A - V_B$ is the third side of the isosceles triangle formed by their tips and the origin. By the cosine rule,

V_{AB}^2 = 230^2 + 230^2 - 2(230)(230)\cos 120^\circ = 230^2(2 + 1) = 3 \times 230^2

so $V_{AB} = 230\sqrt{3} = 398\,\mathrm{V} \approx 400\,\mathrm{V}$ . This is why REG quotes the low-voltage network as 400 V/230 V.

Application 3: tuning a radio

Every broadcasting station in range induces a tiny alternating emf in a receiver's aerial, each at its own carrier frequency. The receiver must pick out one. It does so with a coil and a variable capacitor: the aerial current induces an emf in the coil $L$ , which acts as a small source in series with $L$ and $C$ . At the circuit's resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$ , the current is at its maximum and the voltage across the capacitor is $Q$ times the induced emf (Lesson 3), while stations at other frequencies give a much smaller response. Turning the tuning knob changes $C$ , and so $f_{0}$ , until it matches the wanted station. A high $Q$ (a coil of low resistance) separates stations that are close in frequency.

Figure 4.4: The tuning circuit of a simple radio receiver.

Worked example 5: tuning across the FM band

The FM broadcast band runs from 88 MHz to 108 MHz. A receiver's tuned circuit has $L = 0.10\,\mathrm{\mu H}$ . What range of capacitance must its variable capacitor cover?

Solution

From $f_0 = \frac{1}{2\pi\sqrt{LC}}$ , $C = \frac{1}{4\pi^2f_0^2L}$ .

At 88 MHz: $C = \frac{1}{4\pi^2 \times (88 \times 10^{6})^2 \times 1.0 \times 10^{-7}} = 3.27 \times 10^{-11}\,\mathrm{F} = 32.7\,\mathrm{pF}$ .

At 108 MHz: $C = 21.7\,\mathrm{pF}$ . So the capacitor must vary from about 22 pF to 33 pF. The HIGHEST frequency needs the SMALLEST capacitance, because $f_0 \propto \frac{1}{\sqrt{C}}$ .

Warning

Common mistakes. (1) Writing $P = V_{\text{rms}}I_{\text{rms}}$ for a reactive circuit: that is the apparent power in VA; multiply by $\cos\phi$ . (2) Using peak values in $P = VI\cos\phi$ without the factor $\frac{1}{2}$ . (3) Connecting power-factor capacitors in SERIES with the load: they go in parallel, so the load's voltage is unchanged. (4) Adding phase voltages to get the line voltage: 230 + 230 = 460 V is wrong; the phasor difference gives $230\sqrt{3}$ .

Tip

Draw the power triangle for every power question: $P$ along, $Q$ up (lagging) or down (leading), $S$ on the hypotenuse, angle $\phi$ at the corner. Then $\cos\phi = \frac{P}{S}$ , $\tan\phi = \frac{Q}{P}$ , and $S^{2} = P^{2} + Q^{2}$ all come straight off the picture.

Real vs Reactive Power (Physics Videos by Eugene Khutoryansky). Watch how energy flows back and forth between the source and a reactive load while only the in-phase part of the current delivers net power.

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Practice

A load takes 6.0 A at 230 V rms with the current lagging by 40°. Find the apparent power, the true power, the reactive power and the power factor.
Show, from $p = v i$ , that a pure capacitor takes no average power.
A series circuit has $R = 30\,\mathrm{\Omega}$ and $Z = 50\,\mathrm{\Omega}$ and carries 2.0 A. Find the true power by two different methods.
A four-pole induction motor on a 50 Hz supply runs with 3.0% slip. Find its speed in rev/min.
A factory load takes 40 kW at power factor 0.80 lagging from a 400 V, 50 Hz supply (single-line model). Find the capacitance needed in parallel to raise the power factor to 1.0.
A radio's tuned circuit has a 250 µH coil. Find the capacitance needed to receive a medium-wave station at 1.0 MHz.
Explain why a supplier may charge a large customer for kVA (apparent power) as well as kWh (energy).
Umutoni says: "A power factor of 0.5 means half of the energy supplied is wasted." Evaluate her statement.

Show answersHide answers

$S = 230 \times 6.0 = 1380\,\mathrm{VA}$ ; $P = 1380\cos 40^\circ = 1060\,\mathrm{W}$ (1057 W); $Q = 1380\sin 40^\circ = 887\,\mathrm{var}$ ; power factor 0.77 lagging.
For a capacitor the current leads by 90°: with $v = V_0\sin\omega t$ , $i = I_0\cos\omega t$ , so $p = \frac{1}{2}V_0I_0\sin 2\omega t$ , whose mean over a cycle is zero.
$P = I^2R = 4.0 \times 30 = 120\,\mathrm{W}$ . Or $V = IZ = 100\,\mathrm{V}$ , $\cos\phi = \frac{30}{50} = 0.60$ , $P = 100 \times 2.0 \times 0.60 = 120\,\mathrm{W}$ .
$n_s = 1500\,\mathrm{rev/min}$ ; $n = 1500 \times (1 - 0.030) = 1455\,\mathrm{rev/min}$ .
$Q = P\tan(\cos^{-1}0.80) = 40 \times 0.75 = 30\,\mathrm{kvar}$ , all to be cancelled. $C = \frac{Q}{\omega V^2} = \frac{30000}{2\pi \times 50 \times 400^2} = 5.97 \times 10^{-4}\,\mathrm{F} \approx 600\,\mathrm{\mu F}$ .
$C = \frac{1}{4\pi^2f^2L} = \frac{1}{4\pi^2 \times 10^{12} \times 2.5 \times 10^{-4}} = 1.0 \times 10^{-10}\,\mathrm{F} = 101\,\mathrm{pF}$ .
The cables, transformers and generators must be sized for the full current, which depends on the apparent power $S = V I$ , not on the true power. A low power factor makes the supplier carry and lose ( $I^{2} R$ ) more current for the same energy sold, so a kVA charge passes that cost to the customer and rewards correction.
Wrong. The reactive part of the energy is not wasted: it flows to the load and back each cycle. The true power $P = S\cos\phi$ is delivered usefully. What IS wasted is the extra $I^{2} R$ heating in the supply lines caused by the larger current, which is why a low power factor is undesirable.

Key points

$P = V_{\text{rms}}I_{\text{rms}}\cos\phi$ ; $S = V_{\text{rms}}I_{\text{rms}}$ (VA); $Q = V_{\text{rms}}I_{\text{rms}}\sin\phi$ (var); power factor $\cos\phi = \frac{P}{S} = \frac{R}{Z}$ .
Parallel capacitors correct a lagging power factor, cutting the current and the line losses $\frac{P^2R}{V^2\cos^2\phi}$ .
Three phases 120° apart give a rotating field ( $n_s = \frac{f}{p}$ ) and a line voltage $\sqrt{3} \times 230 \approx 400\,\mathrm{V}$ .
A radio tunes by resonance: $f_0 = \frac{1}{2\pi\sqrt{LC}}$ , adjusted with a variable capacitor.

1Why the grid cares about the angle

2The average power in an AC circuit

3Power factor correction

4Application 1: transmitting power

5Application 2: AC motors

6Application 3: tuning a radio

Why the grid cares about the angle

The average power in an AC circuit

Power factor correction

Application 1: transmitting power

Application 2: AC motors

Application 3: tuning a radio