Power in A.C Circuits

Chapter 26: Power in A.C Circuits

Definitions

Power is the rate of flow of energy past a given point in the circuit.[cite: 622] In AC, inductors and capacitors may result in the periodic reversal of the direction of energy flow.[cite: 623]

Real Power

The portion of power averaged over a complete cycle of the AC waveform results in a net transfer of energy in one direction; this is known as the active power.[cite: 625, 626, 627, 628]

Reactive Power

The portion of stored power due to stored energy, which returns to the source in each cycle, is known as the reactive power.[cite: 632, 633, 634]

Power factor is important to consider because a power factor less than 1 means an AC circuit’s wiring has to carry more current than would be necessary with zero reactance, to deliver the same amount of true power to the resistive load.[cite: 647, 648, 649, 650, 651, 652]

26.1 Introduction

The instantaneous power is the product of the changing voltage v and changing current i.[cite: 661, 662] Therefore, their product is given in Watts:[cite: 664]

p = vi WATTS[cite: 664]

26.2 Determination of Power in AC Circuits

a) Purely Resistive AC Circuit

Let v = V_m \sin \omega t.[cite: 667] The resulting current is i = I_m \sin \omega t.[cite: 669, 670] The corresponding power is:[cite: 671]

p = vi = V_m \sin \omega t \times I_m \sin \omega t = V_mI_m \sin² \omega t[cite: 674, 676]

From the double angle formula \cos 2A = 1 – 2\sin² A, we get \sin² \omega t = \frac{1}{2}(1 – \cos 2\omega t).[cite: 675, 678, 679] Therefore:[cite: 680]

p = \frac{1}{2}V_mI_m(1 – \cos 2\omega t)[cite: 681]

The average power developed over a cycle is:[cite: 690]

P = \frac{1}{2}V_mI_m = VI WATTS[cite: 690]
P = I²R = \frac{V²}{R} WATTS[cite: 691, 697]

b) Power in a Purely Inductive AC Circuit

Voltage leads current by 90° in a purely inductive circuit.[cite: 699] If v = V_m \sin \omega t, then i = I_m \sin(\omega t – \pi/2).[cite: 700]

p = vi = V_mI_m \sin \omega t \times \sin(\omega t – \pi/2)[cite: 700]

However, \sin(\omega t – \pi/2) = -\cos \omega t.[cite: 701, 702] Rearranging gives p = -\frac{1}{2}V_mI_m(2 \sin \omega t \cos \omega t).[cite: 703] Using the double angle formula, 2 \sin \omega t \cos \omega t = \sin 2\omega t:[cite: 704]

p = -\frac{1}{2}V_mI_m \sin 2\omega t[cite: 704]

Since the power is sinusoidal, the average power over a complete cycle in a purely inductive circuit is zero.[cite: 707, 708, 723]

c) Purely Capacitive AC Circuit

Current leads voltage by 90° in a purely capacitive circuit.[cite: 728, 729] Let v = V_m \sin \omega t and i = I_m \sin(\omega t + \pi/2).[cite: 727, 730]

p = vi = V_mI_m \sin \omega t \sin(\omega t + \pi/2)[cite: 732, 733]

Since \sin(\omega t + \pi/2) = \cos \omega t:[cite: 734, 735]

p = \frac{1}{2}V_mI_m \sin 2\omega t[cite: 741, 742]

d) R-L / R-C AC Circuit

For an R-L or R-C circuit, let v = V_m \sin \omega t and i = I_m \sin(\omega t + \phi).[cite: 753, 756, 758] The phase angle \phi will be positive for an R-C and negative for an R-L circuit.[cite: 758] The instantaneous power p is:[cite: 759]

p = vi = V_mI_m \sin \omega t \sin(\omega t + \phi)[cite: 760]

Using the rule \sin A \sin B = \frac{1}{2}[\cos(A-B) – \cos(A+B)], substituting A = \omega t and B = \omega t + \phi gives:[cite: 765, 767, 768, 769]

p = \frac{1}{2}V_mI_m[\cos(-\phi) – \cos(2\omega t + \phi)][cite: 776]

However, \cos(-\phi) = \cos(\phi), thus p = \frac{1}{2}V_mI_m[\cos(\phi) – \cos(2\omega t + \phi)].[cite: 777, 778] Since \cos(2\omega t + \phi) has a mean value of 0 over a cycle:[cite: 779, 781, 782]

P = \frac{1}{2}V_mI_m \cos \phi = VI \cos \phi WATTS[cite: 783, 785]

26.3 Power Triangle and Power Factor

The Power Triangle consists of:[cite: 812]

True or Active Power (P): P = VI \cos \phi (Watts)[cite: 813]
Apparent Power (S): S = VI (Volt-Amperes, VA)[cite: 815, 817]
Reactive Power (Q): Q = VI \sin \phi (Volt-Amperes Reactive, var)[cite: 816, 819, 821]

Using a phasor diagram, we get the power triangle: S = P + jQ.[cite: 822, 840, 841]

Why Apparent Power is Important

Transformers, generators, and cables are usually rated in volt-amperes (VA) not WATTS.[cite: 846, 847] The allowable output of these items is usually limited by heat loss, which is determined by the voltage and current, almost independent of the power factor.[cite: 849]

Inductive and Capacitive Reactive Power

Inductive reactive power is defined as positive by convention, and capacitive reactive power is defined as negative by convention.[cite: 871, 872]

Power Factor

Power Factor = \frac{P}{S} = \frac{VI \cos \phi}{VI} = \cos \phi = \frac{R}{Z}[cite: 874]

A circuit in which the current lags voltage (i.e. an inductive circuit) is said to have a lagging power factor and lagging reactive power.[cite: 876, 877, 879, 880, 881] The opposite is true for a capacitive circuit.[cite: 882]

26.4 Use of Complex Numbers for Determination of Power

Let V = V(\cos \alpha + j\sin \alpha) = a + jb and I = I(\cos \beta + j\sin \beta) = c + jd.[cite: 886, 887, 888, 890]

Active Power

The power P = VI \cos \phi where \phi is the angle between the voltage and current, thus P = VI \cos(\alpha – \beta).[cite: 893, 894, 895, 896] From the compound angle formula \cos(\alpha – \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta:[cite: 897, 898]

P = VI[\cos \alpha \cos \beta + \sin \alpha \sin \beta] = ac + bd[cite: 899]

Reactive Power

Reactive power Q = VI \sin(\alpha – \beta).[cite: 904] From the compound angle formula \sin(\alpha – \beta) = \sin \alpha \cos \beta – \cos \alpha \sin \beta:[cite: 905, 906]

Q = VI(\sin \alpha \cos \beta – \cos \alpha \sin \beta) = bc – ad[cite: 908, 911]

26.5 Power Factor Improvement

A high power factor reduces the current flowing, which means lower power losses due to I²R and hence results in cheaper electricity tariffs.[cite: 928, 929, 930, 932]

Most residential and industrial loads are inductive, meaning they operate at a lagging power factor.[cite: 935, 936, 937] In order to reduce the value of S (VA), a capacitor can be connected in parallel with the L-R load.[cite: 937, 938, 939]

The effect of the capacitor is to reduce the reactive power of systems without changing the active power.[cite: 946] The angle is reduced, decreasing the volt-amperes and causing a decrease in current for the same active power, thus causing an increase in efficiency.[cite: 970, 971, 972, 973, 974, 976]

The use of synchronous motors is another method of power factor improvement as these can be made to operate at a leading power factor.[cite: 977, 979, 980, 981]