# Solution

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). 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 3K and 1K are in series, but then that combination is shorted out by the wire that replaced the voltage source. Another way to think about it is that we have 1K+3K =4K, and then 4K || 0 = 0. That is, a 0 ohm resistor in parallel with anything else is still 0. Thus we have:

Redrawing it slightly (but maintaining the same connections):

We now see the 4K is in parallel to the 5K, so

Rth = 4K || 5K = 4K*5K/(4K+5K) = 2.22K ohms

Now we must find Vth. For this, we must find the open circuit voltage at the terminals:

Note that there is 2V across the 4K and 5K in series. It does not matter that the 2V is also across the 3K and 1K in series. We will use a voltage divider for the 4K and 5K in series with a know total voltage of 2V:

Vth = 5K/(4K+5K)*2V = 1.11V

So the final Thevenin equivalent is:

The Norton could be found directly by computing Rth in the same way and finding Isc by shoring the terminals and computing the current. Now that we have the Thevenin, we can also find the Norton simply by Ohm's law: Isc = Vth/Rth = 1.11V/2.22Kohm = 0.5mA. The Norton is then:

Back to Thevenin/Norton examples

© Calvin College, 2003
This page was written and is maintained by Steve VanderLeest.
It was last modified on 10-Apr-2003 .