Question
In the problem below, where did they get the below equation from?
Determine the capacitor current response iC(t) to the rectangular pulse shown in Fig. 6-16(a), given that R= 125 L = 0.1 H, and C = 4 mF. Solution: Pert the proposed solution recipe, our first should be to determine iL(0-) and upsilonC(0-). In the present prior to activating the current source, the circuit contained the energy. Hence, iL(0-) = 0 and upsilonC(0-) = 0, in which car transformation of the circuit elements to the s-domain replacing L with sL and C with l/(sC). The s-domain is shown in Fig. 6-16(b). The input source is is(t) = 6u(t) - 6u(i - 1), (6,113) RLC circuit in time domain s-domain Current waveforms Figure 6-16: Circuit of Example 6-12 and, according to entries #2 and #2a in Table 6-4, the s-domain expression for the source should be Is = 6/s - 6/s e-s. (6.114) Our intermediate goal is to determine IC, the s-domain current through the capacitor in Fig. 6-16(b). Application of KCL at node V in Fig. 6-16(b) gives V(1/R + 1/sL + sC) = I5. (6.115) Also, IC is related to V by IC = sCV. Solution for IC leads to IC = (s2/s2 + s/RC + l/LC) Is. (6.116) where, in anticipation of applying partial-fraction expansion Enter on, we configured the denominator such that the coefficient of the highest-order s-term is 1. Since Is is the sum of two similar terms, we shall apply the superposition principle as follows: IC = (s2/s2 + s/RC + 1/LC) (6/s - 6e-s/s) = IC1 + IC2. (6.117) where IC1 = (s2/s2 + s/RC + 1/LC) 6/s = 6s/s2 + s/RC + 1/LC, (6.118a) and IC2 = (6s/s2 + s/RC + 1/LC) (-e-s). (6.118b)
Explanation / Answer
(6.115)
V ( 1/R + 1/sL + sC ) = Is
Thus, V = Is / (1/R + 1/sL + sC )
Also, Ic = sCV
Thus, putting value of V from 6.115 here:
Ic = sC * Is / (1/R + 1/sL + sC )
Thus,. Ic = (s^2 / (s^2 + s/Rc + 1/LC) ) * Is
