A current i(t)=50e-5 mA is applied to a capacitor C = 1000 F shown in Fig. P9.31. Figure P9.31 i(t) I C V(f) A current applied to a capacitor. (a) Knowing that i(t) = Cd and that the initial voltage is v(0) = 50 V, integrate both sides of the equation to determine the output voltage v(t). (b) Evaluate the voltage v(t) for t = 0.25 s, t = 0.50 s, and 1 = 0.75 s and use your results to plot v(t) for (c) Suppose the voltage across the capacitor is v(t)=10(6-e-5) volts. Compute the power p(t) = v(t)i(t). (d) Suppose that the stored power is P(t)=3e-5-0.5e-10 W. Knowing that p(t) = integrate both sides of the equation and calculate the stored energy. Assume the initial stored energy is zero (i.e., W(0) = OJ). dW dt,