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[CHAP. 6

Fig. 6-20

Let v designate the voltage across the parallel RC combination. The current in R is iR ¼ v=R ¼ 106 v. During the pulse, iR remains negligible because v cannot exceed 1 V and iR remains under 1 mA. Therefore, it is reasonable to assume that during the pulse, iC ¼ 1 A and consequently vð0þ Þ ¼ 1 V. For t > 0, from application of KVL around the RC loop we get vþ

dv ¼ 0; dt

vð0þ Þ ¼ 1 V


The only solution to (41) is v ¼ et for t > 0 or vðtÞ ¼ et uðtÞ for all t. For all practical purposes, is can be considered an impulse of size 106 A, and then v ¼ et uðtÞ (V) is called the response of the RC combination to the current impulse.


Plot the function vðtÞ which varies exponentially from 5 V at t ¼ 0 to 12 V at t ¼ 1 with a time constant of 2 s. Write the equation for vðtÞ. Identify the initial point A (t ¼ 0 and v ¼ 5Þ and the asymptote v ¼ 12 in Fig. 6-21. The tangent at A intersects the asymptote at t ¼ 2, which is point B on the line. Draw the tangent line AB. Identify point C belonging to the curve at t ¼ 2. For a more accurate plot, identify point D at t ¼ 4. Draw the curve as shown. The equation is vðtÞ ¼ Aet=2 þ B. From the initial and final conditions, we get vð0Þ ¼ A þ B ¼ 5 and vð1Þ ¼ B ¼ 12 or A ¼ 7, and vðtÞ ¼ 7et=2 þ 12.

Fig. 6-21


The voltage v ¼ V0 eajtj for a > 0 is connected across a parallel combination of a resistor and a capacitor as shown in Fig. 6-22(a). (a) Find currents iC , iR , and i ¼ iC þ iR . (b) Compute and graph v, iC , iR , and i for V0 ¼ 10 V, C ¼ 1 mF, R ¼ 1 M, and a ¼ 1. (a) See (a) in Table 6-3 for the required currents. (b) See (b) in Table 6-3. Figures 6-22(b)–(e) show the plots of v, iC , iR , and i, respectively, for the given data. During t > 0, i ¼ 0, and the voltage source does not supply any current to the RC combination. The resistor current needed to sustain the exponential voltage across it is supplied by the capacitor.