Ex:           In the circuit below, based on the interpretation that an uncharged capacitance initially looks like a short circuit and finally like an open circuit, give a qualitative description of the current i over time.

 

 

 

Ans:         When the switch is first closed, C looks like a short circuit. Hence at t = 0+, i can be found from voltage division to be 5 mA. As t -> ∞, C begins to look like an open circuit, in which case i can again be found from current division to be i = 3 mA. Thus the asymptotic value of i is 3 mA. Consequently, i decreases smoothly from an initial value of 5 mA to a final value of 3 mA at a rate that depends on the time constant, Rt C2, where Rt is the Thevenin equivalent resistance across C. For a small value of C, the decay will be fast; for a larger value of C, the decay will be slower.