Hint

Keep your eye on the voltage across the capacitor. That is the only voltage/current guaranteed not to change instantaneously.

There are several time periods to consider:

• Before time 0: This is before anything happens. Normally we assume the circuit has been in this situation for a very long time (many, many time constants). That means the capacitor will be acting like an open as we approach time 0. Compute the voltage across the capacitor (the current through it is zero, but the voltage across it is not zero). The current through the 8K resistor at this time is 0.
• At time 0: At this moment, the 2K resistor becomes disconnected from the circuit. Also, the capacitor will not allow its voltage to be changed. Thus the capacitor voltage just after time 0 will be the same as the capacitor voltage just before time 0 (computed in the previous time period).
• Between time 0 and 100ms: The voltage on the capacitor will exponentially decay from its initial value (computed in the previous step) towards its final value. It will not actually reach the final value (because something happens at 100ms), but we need to compute it in order to write the capacitor voltage formula. This is a first order transient, so you'll need to find the Thevenin resistance seen by the capacitor in order to compute the time constant. Remember that the 2K resistor is NOT in the circuit at this time. Use the capacitor voltage to back-track to find the current through the 8K resistor.
• At time 100ms: Compute the voltage across the capacitor using the formula you derived in the previous step. This will be your initial capacitor voltage. Compute the final voltage and the Thevenin resistance seen by the capacitor (remember that the 2K resistor is back in the circuit now). Use the capacitor voltage to back-track to find the current through the 8K resistor.

Back to Transient Analysis Examples

© Steven H. VanderLeest, 2007
This page was written and is maintained by Steve VanderLeest.
It was last modified on 11-May-2007 .