Since we are finding the Thevenin with respect to the capacitor, we also take the cap out of the circuit and consider the resistance seen from the terminals where the cap was.

The Thevenin equivalent has two parts, Vth and Rth. We will do the easier one first -- Rth. To find the Thevenin resistance, deactivate all sources (short voltages and open currents).

From the point of view of the capacitor terminals, the 1K and 2K are shorted out. The 7K also is not included because no current can flow through it. If current was fed into the top terminal, it would flow through the 4K and 5K and then come back through the other terminal. Thus

Rth = 4K || 5K = (4K*5K)/(4K+5K) = 2.2K ohms

Next, we'll find Vth using node-voltage analysis, with one node (the bottom wire is the reference node).

Writing KCL at the node V1 (current leaving):

Solve for V by multiplying through by 20K:

5V - 15 + 4V - 120 = 0

9V = 135

V = 135/9 = 15V

So the final Thevenin Equivalent is:

Back to Thevenin/Norton examples

© Calvin College, 2004

This page was written and is maintained by Steve VanderLeest.

It was last modified on
17-Mar-2004
.