Remove the capacitor, since it is not part of the circuit we wish to simplify.
In order to find the Norton Short-circuit current, short the terminals where the capacitor used to be, since we are finding the Norton
Now the 2k ohm resistor is shorted-out, so we can eliminate it. Then we'll find the current through the short. Here are two different ways to solve for the current:
Now find the Norton resistance (same as Thevenin resistance). First, open the terminals where the capacitor used to be:
Now deactivate all sources (short voltage sources, open current sources):
The 4k ohm is shorted out, so it can be removed, and the 5k cannot have any current through it, so it can also be removed.
We can combine the 8k, 3k, and 1k in series.
However, the 12k combination is shorted out by the wire, so it can be eliminated.
Now let's redraw the circuit:
The resistors are in parallel, so the total resistance seen by the capacitor is
Rth = 6k || 2k = 6k*2k/(6k+2k) = 12M/8k = 1.5k ohms
Thus the final Norton is shown below:
© Calvin College, 2000
This page was written and is maintained by Steve VanderLeest.
It was last modified on 1 Mar 2000.