Superposition Solution for Isc

Superposition analysis to find the current through the short means we solve several simpler circuits (one for each source. Since there are three sources, we have three simpler circuits to solve.

In each of these circuits, all sources but one have been deactivated (voltage sources are shorted, current sources are opened). Our total answer is the sum of the answers we get from each of the circuits:

Isc = Isc15 + Isc12 + Isc20

First, let's combine the 8k, 3k, and 1k in series:

R = 8k + 3k + 1k = 12k

and also drop the 5k, since no current can flow through it.

The 12k is shorted out by the horizontal wire, so it can be removed.

Now we can see that the current through the 6k ohm resistor will be the current Isc. Thus,

Isc15 = 15V/6k = 2.5mA

That's not the final answer though! We have to find the other two parts of Isc.

The 5k ohm is again left hanging, so we can remove it. The 8k, 3k, and 1k can be combined in series (12k). The 4k ohm is shorted out, so it can be removed.

Isc12 = -12V/6k = -2mA

We can remove the 4k and 6k since they are both shorted out. We can also combine the 8k and 3k in series (11k).

Isc20 = 20mA

Now we add the three sub-answers together to get our final answer:

Isc = Isc15 + Isc12 + Isc20

= 2.5mA - 2mA + 20mA

= 20.5 mA

Calvin College, 2000
This page was written and is maintained by Steve VanderLeest.
It was last modified on 1 Mar 2000.