AC Power Analysis

The instantaneous power is the product of the instantaneous voltage and current, exactly as in Chapter 1. Taking

the product is

Applying the identity \(\cos A\cos B = \tfrac{1}{2}\left[\cos(A-B) + \cos(A+B)\right]\):

The result splits into two parts, and the split is the key to everything in this chapter:

• The first term is constant — independent of time. Its value depends only on the phase difference \(\theta_{v}-\theta_{i}\) between voltage and current.

• The second term is sinusoidal at twice the source frequency. Over a complete cycle its average is zero, because the positive half-cycle exactly cancels the negative.

When \(p(t)\) is positive the circuit absorbs power; when it is negative, power flows back to the source. That reversal is possible only because storage elements are present — a resistive circuit never returns energy. The larger the reactive content, the more of each cycle is spent giving energy back.

Section 4.8 showed that a DC source delivers maximum power when \(R_{L} = R_{Th}\). The AC case has an extra degree of freedom, because both the source and the load now have reactance as well as resistance.

With \(\mathbf{Z}_{Th} = R_{Th}+jX_{Th}\) and \(\mathbf{Z}_{L} = R_{L}+jX_{L}\), the current and the power delivered are

This must be maximised with respect to two variables. Setting \(\partial P/\partial X_{L} = 0\) requires the term \(\left(X_{Th}+X_{L}\right)\) to vanish — which makes sense, since that term appears only in the denominator and any non-zero value can only reduce \(P\):

Setting \(\partial P/\partial R_{L} = 0\) then gives

once the first condition is imposed. Combining the two:

Maximum average power is transferred when the load impedance is the complex conjugate of the Thévenin impedance. This is the conjugate match, and it is not the same as \(\mathbf{Z}_{L} = \mathbf{Z}_{Th}\) — a distinction that catches many students.

Physically, the conjugate cancels the source reactance: \(X_{Th}+X_{L} = 0\) makes the total circuit purely resistive, which is resonance. With the reactance gone, the remaining problem is the DC one of Chapter 4, and its answer is \(R_{L}=R_{Th}\) as before. Substituting back:

The factor of 8 rather than the DC formula's 4 arises because \(\left|\mathbf{V}_{Th}\right|\) here is an amplitude, not an RMS value. In terms of RMS quantities the formula reverts to \(V_{Th,\mathrm{rms}}^{2}/4R_{Th}\).

A restricted case. If the load is constrained to be purely resistive — no reactance available — then \(X_{L}=0\) and the optimum becomes

A sinusoid's average value over a full cycle is zero, so quoting it is useless. The meaningful measure of a periodic waveform's "size" is defined by what it does:

The effective value of a periodic current is the DC current that delivers the same average power to a resistor.

Equating the average power of the periodic current with that of the equivalent DC:

The name follows the operations performed, read inside out: take the root of the mean of the square. The effective value and the RMS value are the same thing, and the terms are used interchangeably.

For a sinusoid \(i = I_{m}\cos\omega t\), using \(\cos^{2}\theta = \tfrac{1}{2}\left(1+\cos 2\theta\right)\):

This factor applies to sinusoids only. Any other waveform has its own ratio, and using \(1/\sqrt{2}\) indiscriminately is a common and serious error.

Average power written in RMS terms loses the awkward factor of one half:

This is why power engineering works almost entirely in RMS. When a supply is described as "240 V", that is an RMS value: the peak is \(240\sqrt{2} = 339~\mathrm{V}\), which is what the insulation must withstand.

Writing the average power as

separates it into two factors, and each deserves a name.

• Apparent power \(S = V_{\mathrm{rms}}I_{\mathrm{rms}}\) is what the power appears to be from a casual multiplication of the meter readings. Its unit is the volt-ampere (VA), deliberately distinct from the watt as a reminder that it is not power in the energetic sense.

• Power factor \(\mathrm{pf} = \cos\left(\theta_{v}-\theta_{i}\right) = P/S\) is the dimensionless fraction of apparent power that does useful work. It lies between 0 and 1.

The angle \(\theta_{v}-\theta_{i}\) is called the power factor angle, and it equals the angle of the load impedance, since

Because the cosine is an even function, the power factor alone does not reveal the sign of the angle. The qualifier must always be given:

• Lagging power factor — current lags voltage, \(\theta_{v}-\theta_{i} > 0\), inductive load. Nearly all industrial loads.

• Leading power factor — current leads voltage, \(\theta_{v}-\theta_{i} < 0\), capacitive load.

• Unity power factor — in phase, purely resistive.

"A power factor of 0.8" is an incomplete specification; "0.8 lagging" is a complete one.

All the quantities so far can be packed into a single complex number. Define the complex power as

Expanding it in polar and then rectangular form:

Substituting \(\mathbf{V}_{\mathrm{rms}} = \mathbf{Z}\mathbf{I}_{\mathrm{rms}}\) gives an equally useful form:

so the real part of the complex power comes from the resistance and the imaginary part from the reactance — precisely as one would hope.

Real power \(P\) is the average power in watts actually delivered to and dissipated by the load. It is the only useful power, and the only quantity an energy meter records.

Reactive power \(Q\) measures the energy exchanged back and forth between the source and the reactive part of the load. Its unit is the volt-ampere reactive (VAR) — again deliberately distinct from the watt. Its sign classifies the load:

• \(Q = 0\) for a resistive load — unity power factor.

• \(Q > 0\) for an inductive load — lagging power factor. An inductor is said to absorb reactive power.

• \(Q < 0\) for a capacitive load — leading power factor. A capacitor supplies reactive power.

That last pair of statements is the entire basis of power factor correction: if an inductive load absorbs VARs and a capacitor supplies them, then putting a capacitor next to the load lets them settle the account locally instead of dragging it back to the generator.

Section 1.5 established that power is conserved in a DC circuit. The same holds in AC, and it holds separately for the real and reactive parts:

The complex powers of the individual elements add to the complex power supplied by the source, whether the elements are in series or in parallel. This follows from the linearity of the phasor transform and is a direct consequence of Kirchhoff's laws.

What does not add is apparent power. Since \(S = \left|\mathbf{S}\right|\) is a magnitude, and magnitudes of complex numbers do not add, \(S_{\text{total}} \neq \sum S_{k}\) in general. This is exactly the trap of Section 9.9, in a new guise.

Conservation of complex power is the best available check on an AC circuit solution. Compute \(\mathbf{S}\) for every element, add them, and confirm the sum matches the source. It costs a minute and catches most errors.

Nearly every practical load is inductive. Induction motors, transformers, fluorescent ballasts, welding sets, washing machines, air conditioners and refrigerators all operate at a lagging power factor, often as low as 0.6 or 0.7.

The consequences fall on the supply system rather than the consumer's energy meter. For a given real power \(P\) at a fixed voltage, the current drawn is

so halving the power factor doubles the current. That extra current dissipates \(I^{2}R\) in every cable and transformer between the load and the generator, and it consumes capacity that could have carried useful load. None of it registers on an energy meter.

Power factor correction is the process of increasing the power factor without altering the voltage or current supplied to the original load. The inductive nature of the load cannot be changed — but a capacitor connected in parallel with it can supply the reactive power that the load demands, so that the supply need not.

Design. The real power is unchanged by the capacitor, since an ideal capacitor consumes none. Only the reactive power changes:

The capacitor must supply the difference:

and since a capacitor across the supply voltage provides \(Q_{C} = V_{\mathrm{rms}}^{2}/X_{C} = \omega C V_{\mathrm{rms}}^{2}\),

The capacitor is connected in parallel with the load, not in series, so that the load continues to see the same voltage and behaves exactly as before. From the supply's point of view the two together draw less current at a better angle.

A wattmeter measures average power, and it must therefore respond to the product of instantaneous voltage and current rather than to either alone. The classical instrument that achieves this is the electrodynamometer.

It contains two coils. The current coil is of few turns and low resistance, connected in series with the load so that it carries the load current. The voltage coil — or pressure coil — is of many turns with a high series resistance, connected across the load so that its current is proportional to the load voltage. The deflecting torque is proportional to the product of the two coil currents, and because the moving system has too much inertia to follow a \(2\omega\) variation, the pointer settles at the average of that product:

The instrument reads watts directly, and — unlike a moving-iron or moving-coil meter — it reads correctly on both AC and DC.

Low power factor wattmeters. At very low power factor the deflecting torque becomes small, and a standard instrument would give an unreadably small deflection with poor accuracy. An LPF wattmeter uses a voltage coil of fewer turns and compensating windings to restore usable deflection. Its scale is then calibrated for a nominal power factor, and the true power requires a multiplying factor:

Forgetting the multiplying factor is a classical laboratory error, and it produces a result wrong by a factor of several.

Chapter 12 returns to power measurement for three-phase systems, where the two-wattmeter method allows the total power of a three-wire system to be measured with only two instruments — and, remarkably, allows the power factor to be deduced from their two readings.

Section 1.7 computed a domestic bill from energy alone. Industrial tariffs are different, and the difference is where the economics of power factor becomes concrete.

A domestic consumer pays only for energy, in kWh — a real-power quantity. Power factor is irrelevant to the bill, which is why domestic appliances are not required to correct it.

An industrial consumer typically pays two charges:

• an energy charge per kWh, exactly as domestic; plus

• a demand charge per kVA of maximum demand — an apparent-power quantity, which depends directly on power factor.

The demand charge exists because the utility must size its cables, transformers and switchgear for the current the consumer draws, and that current is set by kVA rather than kW. A consumer at poor power factor demands more of the network's capacity while paying the same energy charge, and the demand charge recovers that cost. Many tariffs go further and impose an explicit penalty below a threshold power factor, or a rebate above it.

• Instantaneous power has a constant part and a part oscillating at \(2\omega\) whose average is zero.

• Average power is \(P = \tfrac{1}{2}V_{m}I_{m}\cos(\theta_{v}-\theta_{i}) = \tfrac{1}{2}\mathrm{Re}\left[\mathbf{V}\mathbf{I}^{*}\right]\). A purely reactive element absorbs none.

• Maximum average power transfer requires the conjugate match \(\mathbf{Z}_{L} = \mathbf{Z}_{Th}^{*}\), giving \(P_{\max} = \left|\mathbf{V}_{Th}\right|^{2}/8R_{Th}\).

• The RMS value is the DC equivalent for power delivery. For a sinusoid it is \(V_{m}/\sqrt{2}\), but that factor applies to sinusoids only.

• Apparent power \(S = V_{\mathrm{rms}}I_{\mathrm{rms}}\) in VA; power factor \(\mathrm{pf} = P/S = \cos\theta\), always qualified as leading or lagging.

• Complex power \(\mathbf{S} = \mathbf{V}_{\mathrm{rms}}\mathbf{I}_{\mathrm{rms}}^{*} = P + jQ\), with \(Q>0\) inductive and \(Q<0\) capacitive.

• Complex power is conserved, so \(P\) and \(Q\) each add separately — but apparent power does not add.

• Power factor correction uses a shunt capacitor supplying \(Q_{C} = P\left(\tan\theta_{1}-\tan\theta_{2}\right)\).

• A wattmeter reads average power; industrial tariffs charge for kVA demand, which is what makes correction economic.

• Omitting the conjugate in \(\mathbf{V}\mathbf{I}^{*}\). Using \(\mathbf{V}\mathbf{I}\) gives the angle \(\theta_{v}+\theta_{i}\) instead of the difference, and hence the wrong \(P\) and \(Q\) entirely.

• Mixing amplitude and RMS forms. \(P = \tfrac{1}{2}V_{m}I_{m}\cos\theta\) uses amplitudes; \(P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\theta\) does not have the half. Using both halves, or neither, gives an error of a factor of two.

• Applying \(V_{m}/\sqrt{2}\) to a non-sinusoidal waveform. That ratio belongs to the sinusoid alone. Use the integral definition for anything else.

• Writing \(\mathbf{Z}_{L} = \mathbf{Z}_{Th}\) for maximum power transfer. The AC condition is the conjugate. The load reactance must be opposite in sign to the source reactance.

• Adding apparent powers. \(S\) is a magnitude and magnitudes do not add. Add the complex powers and take the magnitude at the end.

• Quoting a power factor without "leading" or "lagging". The cosine is even, so 0.8 alone does not determine the angle's sign.

• Confusing the units. Watts for \(P\), VAR for \(Q\), VA for \(S\). They are numerically different quantities and the distinct units exist to prevent exactly this error.

• Connecting the correction capacitor in series. It goes in parallel, so the load continues to see its rated voltage.

• Assuming correction reduces the energy bill. It reduces the demand charge and the distribution losses. The kWh consumed by the load is unchanged.

• Forgetting a wattmeter's multiplying factor on a low-power-factor instrument. The scale reading is not the answer.

Work these before opening the answers. State units precisely — W, VAR or VA — and qualify every power factor.

1. P11.1 A load has \(v = 160\cos\left(\omega t + 20^{\circ}\right)~\mathrm{V}\) and \(i = 4\cos\left(\omega t - 40^{\circ}\right)~\mathrm{A}\). Find the average power.

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   \[P = \tfrac{1}{2}\left(160\right)\left(4\right)\cos\left(20^{\circ}+40^{\circ}\right) = 320\cos 60^{\circ} = 160.0~\mathrm{W}\]

2. P11.2 Find the RMS value of a square wave that alternates between \(+6~\mathrm{A}\) and \(-6~\mathrm{A}\), and its peak factor.

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   The square of the waveform is 36 at all times, so its mean is 36 and

   \[I_{\mathrm{rms}} = \sqrt{36} = 6.000~\mathrm{A}, \qquad \text{peak factor} = \frac{6}{6} = 1.000\]

   A square wave is the only common waveform whose RMS equals its peak.

3. P11.3 A network has \(\mathbf{Z}_{Th} = \left(6+j8\right)~\Omega\) and \(\mathbf{V}_{Th} = 40\angle 0^{\circ}~\mathrm{V}\) (amplitude). Find \(\mathbf{Z}_{L}\) for maximum power transfer and the value of \(P_{\max}\).

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   \[\mathbf{Z}_{L} = \mathbf{Z}_{Th}^{*} = \left(6-j8\right)~\Omega\]

   \[P_{\max} = \frac{40^{2}}{8\left(6\right)} = \frac{1600}{48} = 33.33~\mathrm{W}\]

4. P11.4 For the same network, what purely resistive load gives maximum power, and how much?

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   \[R_{L} = \left|\mathbf{Z}_{Th}\right| = \sqrt{36+64} = 10~\Omega\]

   \[\left|\mathbf{I}\right| = \frac{40}{\left|6+j8+10\right|} = \frac{40}{\left|16+j8\right|} = \frac{40}{17.89} = 2.236~\mathrm{A}\]

   \[P = \tfrac{1}{2}\left(2.236\right)^{2}\left(10\right) = 25.00~\mathrm{W}\]

   Less than the 33.33 W of the conjugate match, as expected.

5. P11.5 A load draws \(\mathbf{S} = \left(1200 + j1600\right)~\mathrm{VA}\). Find \(S\), the power factor, and state whether the load is inductive or capacitive.

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   \[S = \sqrt{1200^{2}+1600^{2}} = 2000~\mathrm{VA}, \qquad \mathrm{pf} = \frac{1200}{2000} = 0.6000\ \text{lagging}\]

   Since \(Q > 0\) the load is inductive.

6. P11.6 A 230 V rms, 50 Hz supply feeds a 3 kW load at 0.7 pf lagging. Find the supply current, and the current after correction to 0.9 lagging.

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   \[I_{1} = \frac{3000}{\left(230\right)\left(0.7\right)} = 18.63~\mathrm{A}, \qquad I_{2} = \frac{3000}{\left(230\right)\left(0.9\right)} = 14.49~\mathrm{A}\]

   A reduction of 22.22 %.

7. P11.7 Find the capacitance required in P11.6.

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   \[\theta_{1} = \cos^{-1}0.7 = 45.57^{\circ}, \quad \tan\theta_{1} = 1.0202; \qquad \theta_{2} = \cos^{-1}0.9 = 25.84^{\circ}, \quad \tan\theta_{2} = 0.4843\]

   \[Q_{C} = 3000\left(1.0202 - 0.4843\right) = 1607.7~\mathrm{VAR}\]

   \[C = \frac{1607.7}{\left(314.2\right)\left(230\right)^{2}} = 96.72~\mu\mathrm{F}\]

8. P11.8 A source supplies \(\left(800+j600\right)~\mathrm{VA}\). One load absorbs \(\left(500+j400\right)~\mathrm{VA}\). Find the complex power of the second load, and check that apparent powers do not add.

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   \[\mathbf{S}_{2} = \left(800+j600\right)-\left(500+j400\right) = \left(300+j200\right)~\mathrm{VA}\]

   Apparent powers: \(\left|\mathbf{S}_{s}\right| = 1000\), \(\left|\mathbf{S}_{1}\right| = 640.3\), \(\left|\mathbf{S}_{2}\right| = 360.6\). Since \(640.3+360.6 = 1000.9 \neq 1000\), the apparent powers indeed do not add — though they come close here because both loads have similar angles.

9. P11.9 A factory draws 800 kW at 0.7 pf. At ₹250 per kVA of demand, what is the monthly saving if corrected to 0.92?

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   \[S_{1} = \frac{800}{0.7} = 1142.9~\mathrm{kVA}, \qquad S_{2} = \frac{800}{0.92} = 869.6~\mathrm{kVA}\]

   \[\text{Saving} = \left(1142.9-869.6\right)\left(250\right) = \text{Rs.}\,68\,320\ \text{per month}\]

10. P11.10 An inductor of reactance \(j20~\Omega\) carries 5 A rms. Find its real and reactive power.

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    \[\mathbf{S} = I_{\mathrm{rms}}^{2}\mathbf{Z} = \left(25\right)\left(j20\right) = j500~\mathrm{VA}\]

    So \(P = 0~\mathrm{W}\) and \(Q = 500~\mathrm{VAR}\). The inductor consumes no energy yet demands 500 VAR from the supply — the central paradox of this chapter, in one line.