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 (make their value 0).

Voltage sources of value 0 can be modeled as shorts. Current sources of value 0 can be modeled as opens.

If current was fed into the top terminal, it would split left and right. The current going left would flow through the 4K (skipping the 1K and 2K because they are shorted out), then split again through the 8K and 9K. The current going right would flow through the 5K (skipping the 7k because of the open).

Rth = 5K || (4K + 8K || 9K) = 3.1K ohms


Next, we'll find Vth using superposition. There are three sources, so we'll have three subproblems to solve. The total Vth can be found as the sum of the individual effects:

Subproblem 1

Vth due to the 3V can be found with this circuit:

Use a voltage divider to find Vth:

Vth = 5K/(4K + 5K + 8K||9K) * 3V = 1.13V

Subproblem 2

Vth due to the 10V can be found with this circuit:

We can find the voltage across the 8K (on our way to finding the voltage across the terminals) by combining the 4K and 5K in series and that in parallel with the 8K:

Req = (5K+4K)||8K = 4.24K

V1 across Req can be found by a voltage divider:

V1 = Req/(Req+9K)*10V = 3.20V

Then the voltage over the 5K (which is also the voltage over the terminals) is another voltage divider on the 3.2V:

Vth = 5K/(4K+5K)*3.2 = 1.78V

Subproblem 3

Vth due to the 6mA can be found with this circuit:

Now combine the 8K and 9K in parallel and that in series with the 4K:

Req = 4K + 8K||9K = 8.24K

Use a current divider to find the current through the 5K branch:

I = Req / (Req + 5K)*6mA = 3.73mA

Use Ohm's law to find the voltage over the 5K:

Vth = 3.73mA * 5K = 18.67V

Now add up the subanswers to get the total Vth:

Vth = 1.13 + 1.78 + 18.67 = 21.58V

The Thevenin equivalent circuit is then:


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 .