Question 7.18: In the circuit in Fig. 7.67, determine the response v(t).

There are two ways of solving this problem using PSpice.
METHOD 1 One way is to first do the dc PSpice analysis to determine the initial capacitor voltage. The schematic of the revelant circuit is in Fig. 7.68(a). Two pseudocomponent VIEWPOINTs are inserted to measure the voltages at nodes 1 and 2. When the circuit is simulated, we obtain the displayed values in Fig. 7.68(a) as V_{1}=0 V \text { and } V_{2}=8 V. Thus the initial capacitor voltage is v(0)=V_{1}-V_{2}=-8 V. The PSpice transient analysis uses this value along with the schematic in Fig. 7.68 is(b). Once the circuit in Fig. 7.68(b) is drawn, we insert the capacitor initial voltage as IC = −8. We select Analysis/Setup/Transient and set Print Step to 0.1 sand Final Step to 4τ = 4 s. After saving the circuit, we select Analysis/Simulate to simulate the circuit. In the Probe menu, we select Trace/Add and display V(R2:2) – V(R3:2) or V(C1:1) – V(C1:2) as the capacitor voltage v(t). The plot of v(t) is shown in Fig. 7.69. This agrees with the result obtained by hand calculation, v(t)=10-18 e^{-t}.

METHOD 2 We can simulate the circuit in Fig. 7.67 directly, since PSpice can handle the open and close switches and determine the initial conditions automatically. Using this approach, the schematic is drawn as shown in Fig. 7.70. After drawing the circuit, we select Analysis/ Setup/Transient and set Print Step to 0.1 s and Final Step to 4τ = 4 s. We save the circuit, then select Analysis/Simulate to simulate the circuit. In the Probe menu, we select Trace/Add and display V(R2:2) – V(R3:2) as the capacitor voltage v(t). The plot of v(t) is the same as that shown in Fig. 7.69.