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CHAPTER 8 Reactance and Impedance
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Problem 120 Repeat problem 118 for the value of generator current, this time using the method of loop currents.
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We begin with the basic case of a capacitor in series with a resistance, to which a sine wave of peak voltage Vp is applied, as in Fig. 136.
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Fig. 136
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In the gure, C is a capacitor of C farads and R is resistance in ohms. Also (as in Fig. 127) v and i denote instantaneous values of voltage and current. Again, as in Fig. 127, we ll be interested only in the permanent SINUSOIDAL STEADY STATE. In this case, the author feels that it s best not to get sidetracked by too many details; hence, let s commence by stating (without bothering about details to begin with) that, corresponding with eq. (194) in section 8.1, the steady-state VOLTAGE EQUATION for Fig. 136 is   1 Vp sin !t RIp sin !t  I sin !t  908 !C p 210 voltage drop voltage drop across C across R in which the applied voltage, Vp sin !t, is taken as the reference wave. Now let s compare eq. (210) with eq. (194), as follows.  phase angle of current, with respect to Vp sin !t, in both equations.) Note, rst, that the angular quantities  and 908 have OPPOSITE SIGNS from what they have in eq. (194). This is due to the fact that the CURRENT in a capacitor LEADS THE VOLTAGE across the capacitor by 908. This happens because FIRST, the amount of electric charge q stored in a capacitor is proportional to capacitor voltage (q vC), and SECOND, capacitor current is the rate of ow of charge (coulombs per second). Now consider the following. In Fig. 128 (section 8.1), suppose the voltage drop were across a capacitor instead of an inductor. Then, since q vC, the curve of charge q would be drawn exactly in phase with the voltage curve. If these changes were made in Fig. 128 it would then be apparent that the curve of the rate of change of q (the current) would lead the curve of voltage drop by 908. Thus, in Fig. 136 we have that the current LEADS the voltage drop across C by 908 but is, as always, IN PHASE with the voltage drop across R. It thus follows that the current wave will lead the applied voltage wave by some intermediate angle , where  will be an
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CHAPTER 8 Reactance and Impedance
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angle between 08 and 908. The actual value of  will depend upon the relative values of R, C, and ! in any given case. Next, from inspection of eq. (210), RIp peak voltage drop across R and 1=!C Ip peak voltage drop across C Now recall, from section 5.8, that the peak value of the SUM of two sinusoidal waves of the same frequency is equal to the vector sum of the peak values of the individual sinusoids. Thus, the relationships of the peak values of the voltages in eq. (210) can be stated in the vector polar form, Vp =08 RIp =8 1=!C Ip =  90 8 211
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which shows that, geometrically, the two peak voltages on the right-hand side can be considered to be the adjacent and opposite sides of a right-angled triangle having Vp as the hypotenuse with  the angle between RIp and Vp (as will be shown later in connection with Fig. 138). Thus, we have that the steady-state PEAK CURRENT is equal to q 212 Ip Vp = R2 1=!C 2 where also  arctan 1=!RC 213
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Next, in Fig. 136 the voltage drop across R is, as always, in phase with the current; thus, since the peak voltage drop across R is equal to RIp , and since the current LEADS the applied voltage Vp by an angle , we have the vector diagram as in Fig. 137.
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