Question 8.10: The input voltage is the unit-step function, u(t). Assume th...

The excitation and the impedance, in the Laplace transform domain, are
V(s)=\frac{1}{s}   and  Z(s)=R+(1/Cs ).
Therefore,
I_{C}(s)=\frac{V(s)}{Z(s)}=\frac{1}{s(R + (1/Cs))} =\frac{1}{R} \left(\frac{1}{(s + (1/RC)} \right).
Applying initial and final value theorems, we get
i(0^{+})=\frac{1}{R} and \underset{t\rightarrow \infty }{\lim } i(t)=0.
With R = 2, 1/R = 1/2 = 0.5. Taking the inverse Laplace transform, we get
i_{C}(t)= \frac{1}{R}e^{-\frac{1}{RC} })u(t).
With With Z_{C} = 1/Cs,
V_{C}(s)=I(s)Z_{C}(s)=\frac{1}{RCs(s + (1/RC))}= \left(\frac{1}{s}-\frac{1}{(s + 1/RC)} \right)  and  v_{C}(t)= (1-e^{-\frac{1}{RC} } )u(t).
The unit-step response is
v_{C}(t)= (1-e^{-\frac{1}{RC} } )u(t). and i_{C}(t)= \frac{1}{R}e^{-\frac{1}{RC} })u(t).
The derivative of these expressions, the unit-impulse response is
v_{C}(t)= \frac{1}{RC}e^{-\frac{1}{RC} }u(t) and i_{C}(t)= -\frac{1}{R^{2} C}e^{-\frac{1}{RC} }u(t).
Initially the impedance of the capacitor is zero and the current is limited by the resistor. In steady state, the excitation becomes DC and the current in the circuit becomes 0 as the impedance of the capacitor is infinite. The excitation voltage appears across the capacitor. With the source connected,the capacitor stores energy 0.5Cv²(t) and the resistor dissipates energy at the rate of Ri²(t). When the source disconnected, the stored energy is eventually dissipated by the resistor. The rate of energy dissipation, which is controlled by the product RC, decreases with the time and finally both the energy and the current become zero. The larger value of RC, the smaller will be the rate of decay. The time constant for the circuit is RC. Figure 8.16a, b shows, respectively, the voltage across the capacitor and the current through for the unit-step source voltage. Figure 8.17a, b shows, respectively, the voltage across the capacitor and the current through for the unit-impulse source voltage.