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v vR vC V sin !t  q and  tan 1 1=!CR V I R2 1=!C 2
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The negative phase angle shifts v to the right of the current i. Consequently i leads v for a series RC circuit. The phase angle is constrained to the range 08  908. For 1=!C ( R, the angle  08, and for 1=!C ) R, the angle  908. See Fig. 9-4.
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Fig. 9-4
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SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
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[CHAP. 9
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PHASORS
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A brief look at the voltage and current sinusoids in the preceding examples shows that the amplitudes and phase di erences are the two principal concerns. A directed line segment, or phasor, such as that shown rotating in a counterclockwise direction at a constant angular velocity ! (rad/s) in Fig. 9-5, has a projection on the horizontal which is a cosine function. The length of the phasor or its magnitude is the amplitude or maximum value of the cosine function. The angle between two positions of the phasor is the phase di erence between the corresponding points on the cosine function.
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Fig. 9-5
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Throughout this book phasors will be de ned from the cosine function. If a voltage or current is expressed as a sine, it will be changed to a cosine by subtracting 908 from the phase. Consider the examples shown in Table 9-2. Observe that the phasors, which are directed line segments and vectorial in nature, are indicated by boldface capitals, for example, V, I. The phase angle of the cosine function is the angle on the phasor. The phasor diagrams here and all that follow may be considered as a snapshot of the counterclockwise-rotating directed line segment taken at t 0. The frequency f (Hz) and ! (rad/s) generally do not appear but they should be kept in mind, since they are implicit in any sinusoidal steady-state problem.
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EXAMPLE 9.3 A series combination of R 10  and L 20 mH has a current i 5:0 cos 500t 108) (A). Obtain the voltages v and V, the phasor current I and sketch the phasor diagram. Using the methods of Example 9.1, vR 50:0 cos 500t 108 di 50:0 cos 500t 1008 dt v vR vL 70:7 cos 500t 558 V vL L
The corresponding phasors are I 5:0 108 A and V 70:7 558 V
CHAP. 9]
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
Table 9-2 Function Phasor Representation
v 150 cos 500t 458
i 3:0 sin 2000t 308 mA 3:0 cos 2000t 608 mA
The phase angle of 458 can be seen in the time-domain graphs of i and v shown in Fig. 9-6(a), and the phasor diagram with I and V shown in Fig. 9-6(b).
Fig. 9-6
Phasors can be treated as complex numbers. When the horizontal axis is identi ed as the real axis of a complex plane, the phasors become complex numbers and the usual rules apply. In view of Euler s identity, there are three equivalent notations for a phasor. polar form rectangular form exponential form The cosine expression may also be written as v V cos !t  Re Ve j !t  Re Ve j!t The exponential form suggests how to treat the product and quotient of phasors. V1 e j1 V2 e j2 V1 V2 e j 1 2 , V1 1 V2 2 V1 V2 1 2 Since V V  V V cos  j sin  V Ve j
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
[CHAP. 9
and, since V1 e j1 = V2 e j2 V1 =V2 e j 1 2 ; V1 V2 1  V1 =V2 1 2
The rectangular form is used in summing or subtracting phasors.
EXAMPLE 9.4 Given V1 25:0 V1 =V2 143:138 and V2 11:2 26:578, nd the ratio V1 =V2 and the sum V1 V2 .
25:0 143:138 2:23 116:568 1:00 j1:99 11:2 26:578 V1 V2 20:0 j15:0 10:0 j5:0 10:0 j20:0 23:36
116:578
IMPEDANCE AND ADMITTANCE
A sinusoidal voltage or current applied to a passive RLC circuit produces a sinusoidal response. With time functions, such as v t and i t , the circuit is said to be in the time domain, Fig. 9-7(a); and when the circuit is analyzed using phasors, it is said to be in the frequency domain, Fig. 9-7(b). The voltage and current may be written, respectively, v t V cos !t  Re Ve j!t i t I cos !t  Re Ie
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