Problem 8.4-1

We will take a capacitor of 0.005 F and not 0.05 F in order to find the same results as the answer given.

First we need to determine v(0), the capacitor voltage at t = 0.

When the switch is open, the capacitor at a steady state acts like an open circuit. Thus the capacitor current is zero and we get v(0) from the only voltage source. We have v(0) = 10 V

When the switch closes we have another 8Ω resistor, and we can find an equivalent Thevenin circuit.

Here R_th = 8*8/16 = 4 Ω

And V_oc = 8*Vs /(8 + 8) = 5 V

We have Vc(t) = V_oc + (v(0) – V_oc)*exp( - t/R_th*C)

Thus Vc(t) = 5+ 5*exp( - 0.5t) for 0 < t < 1.5 s

We have for t = 1.5 s, Vc(1.5) = 5+ 5*exp( -0.5*1.5)

Thus Vc(1.5) = 7.36 V

We can now have the new voltage, once the switch opens again, given by:

Vc(t) = V_oc’ + (Vc(1.5) – V_oc’)*exp( - (t – 1.5)/R_th’*C)

Here, when the switch opens, we have V_oc’ = 10 V

And R_th’ = 8Ω

Vc(1.5) – V_oc’ = -2.64

Finally, for 1.5 < t we have Vc(t) = 10 – 2.64*exp( - 0.25*(t – 1.5))