p(t) dt

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0 T 0

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1 = T Pav =

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VI 1 cos( ) dt + 2 T

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(74) VI cos(2 t ) dt 2 (75)

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VI cos( ) Average power 2 since the second integral is equal to zero and cos( ) is a constant

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Instantaneous and average power 1 Instantaneous power 08 Average power 06 04 02 0 02 04 0 002 004 006 008 01 Time (s)

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Figure 73 Instantaneous and average power dissipation corresponding to the signals plotted in Figure 72

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As shown in Figure 71, the same analysis carried out in equations 71 to 73 can also be repeated using phasor analysis In phasor notation, the current and voltage of equation 71 are given by V(j ) = V ej 0 I(j ) = I e j (76)

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Note further that the impedance of the circuit element shown in Figure 71 is de ned by the phasor voltage and current of equation 76 to be Z= V j ( ) e = |Z|ej Z I (77)

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Watts

7

AC Power

and therefore that the phase angle of the impedance is Z = (78)

The expression for the average power obtained in equation 74 can therefore also be represented using phasor notation, as follows: 1 V2 1 cos = I 2 |Z| cos 2 |Z| 2

Pav =

(79)

AC Power Notation It has already been noted that AC power systems operate at a xed frequency; in North America, this frequency is 60 cycles per second (Hz), corresponding to a radian frequency = 2 60 = 377 rad/s AC power frequency (710)

In Europe and most other parts of the world, AC power is generated at a frequency of 50 Hz (this is the reason why some appliances will not operate under one of the two systems)

It will therefore be understood that for the remainder of this chapter the radian frequency, , is xed at 377 rad/s

With knowledge of the radian frequency of all voltages and currents, it will always be possible to compute the exact magnitude and phase of any impedance in a circuit A second point concerning notation is related to the factor 1 in equation 79 2 It is customary in AC power analysis to employ the rms value of the AC voltages and currents in the circuit (see Section 42) Use of the rms value eliminates the factor 1 in power expressions and leads to considerable simpli cation Thus, the 2 following expressions will be used in this chapter: V Vrms = = V 2 I Irms = = I 2 Pav = V2 1 V2 cos = cos 2 |Z| |Z| 1 2 I |Z| cos = I 2 |Z| cos = V I cos 2 (711) (712) (713)

Part I

Circuits

Figure 74 illustrates the so-called impedance triangle, which provides a convenient graphical interpretation of impedance as a vector in the complex plane From the gure, it is simple to verify that R = |Z| cos X = |Z| sin (714) (715)

R VS

+ ~

Finally, the amplitudes of phasor voltages and currents will be denoted throughout this chapter by means of the rms amplitude We therefore introduce a slight modi cation in the phasor notation of 4 by de ning the following rms phasor quantities: V = Vrms ej V = V ej V = V V and I = Irms ej I = I ej I = I I In other words, (717) (716)

Z R

Figure 74 Impedance triangle

throughout the remainder of this chapter the symbols V and I will denote the rms value of a voltage or a current, and the symbols V and I will denote rms phasor voltages and currents

Also recall the use of the symbol to represent the complex exponential Thus, the sinusoidal waveform corresponding to the phasor current I = I I corresponds to the time-domain waveform (718) i(t) = 2I cos( t + I ) and the sinusoidal form of the phasor voltage V = V V is v(t) = 2V cos( t + V )

(719)