ssrs barcodelib Sinusoidal SteadyState Circuit Analysis in Software

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Sinusoidal SteadyState Circuit Analysis
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9.1 INTRODUCTION This chapter will concentrate on the steady-state response of circuits driven by sinusoidal sources. The response will also be sinusoidal. For a linear circuit, the assumption of a sinusoidal source represents no real restriction, since a source that can be described by a periodic function can be replaced by an equivalent combination (Fourier series) of sinusoids. This matter will be treated in 17.
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ELEMENT RESPONSES
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The voltage-current relationships for the single elements R, L, and C were examined in 2 and summarized in Table 2-1. In this chapter, the functions of v and i will be sines or cosines with the argument !t. ! is the angular frequency and has the unit rad/s. Also, ! 2f , where f is the frequency with unit cycle/s, or more commonly hertz (Hz). Consider an inductance L with i I cos !t 458 A [see Fig. 9-1(a)]. The voltage is vL L di !LI sin !t 458 !LI cos !t 1358 dt V
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Fig. 9-1
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SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
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A comparison of vL and i shows that the current lags the voltage by 908 or =2 rad. The functions are sketched in Fig. 9-1(b). Note that the current function i is to the right of v, and since the horizontal scale is !t, events displaced to the right occur later in time. This illustrates that i lags v. The horizontal scale is in radians, but note that it is also marked in degrees ( 1358; 1808, etc.). This is a case of mixed units just as with !t 458. It is not mathematically correct but is the accepted practice in circuit analysis. The vertical scale indicates two di erent quantities, that is, v and i, so there should be two scales rather than one. While examining this sketch, it is a good time to point out that a sinusoid is completely de ned when its magnitude V or I , frequency (! or f ), and phase (458 or 1358) are speci ed. In Table 9-1 the responses of the three basic circuit elements are shown for applied current i I cos !t and voltage v V cos !t. If sketches are made of these responses, they will show that for a resistance R, v and i are in phase. For an inductance L, i lags v by 908 or =2 rad. And for a capacitance C, i leads v by 908 or =2 rad.
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Table 9-1 i I cos !t v V cos !t
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vr RI cos !t
iR
V cos !t R
vL !LI cos !t 908
iL
V cos !t 908 !L
vC
I cos !t 908 !C
iC !CV cos !t 908
EXAMPLE 9.1 The RL series circuit shown in Fig. 9-2 has a current i I sin !t. Obtain the voltage v across the two circuit elements and sketch v and i. vR RI sin !t di !LI sin !t 908 dt v vR vL RI sin !t !LI sin !t 908 vL L
Fig. 9-2 Since the current is a sine function and v V sin !t  V sin !t cos  V cos !t sin  we have from the above v RI sin !t !LI sin !t cos 908 !LI cos !t sin 908 2 1
CHAP. 9]
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
Equating coe cients of like terms in (1) and (2), V sin  !LI and V cos  RI q v I R2 !L 2 sin !t arctan !L=R q !L V I R2 !L 2 and  tan 1 R
Then and
The functions i and v are sketched in Fig. 9-3. The phase angle , the angle by which i lags v, lies within the range 08  908, with the limiting values attained for !L ( R and !L ) R, respectively. If the circuit had an applied voltage v V sin !t, the resulting current would be V i q sin !t  R2 !L 2 where, as before,  tan 1 !L=R .
Fig. 9-3
EXAMPLE 9.2 If the current driving a series RC circuit is given by i I sin !t, obtain the total voltage across the two elements. vR RI sin !t vC 1=!C sin !t 908
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