NORTON'S THEOREM

In the analysis of electric networks, a theorem stating that combinations of two-terminal AC voltage sources and impedances can be represented by a single current source and a parallel impedance. A DC source would be represented by a single current source and a parallel resistor.

STEPS TO SOLVE THE PROBLEMS

The Norton is very similar to the Thevenin. In fact, you can find the Thevenin first, then just compute the Norton Current (Isc) by using Ohm's law:
Isc = Vth/Rth
You can also find it directly:

1. Find the Norton Current, also known as the short circuit current, by shorting the terminals (place a wire across them) and compute the current through the short using your favorite analysis method (node-voltage, superposition, or whatever).
2. Find the Norton Resistance the same way as the Thevenin Resistance.

QUES1


Find the Norton Equivalent with respect to the 20uF capacitor.

Solution

Remove the capacitor, since it is not part of the circuit we wish to simplify.

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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

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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

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Thus the final Norton is shown below:



QUES2

Find the Norton Equivalent with respect to the 3 Kohm resistor in the middle of the circuit, i.e., the 3 Kohm resistor itself should not be part of the equivalent that you compute.



TRY YOURSELF


HINT:Since we are finding the Norton with respect to the 3 Kohm, we take the 3 Kohm out of the circuit and consider the resistance seen from the terminals where the 3K was.
The circuit to the left of the 3K is already a Norton equivalent, where the Norton current is -10 mA (because it is facing down). The resistance is infinite. That is, when you open the current source to deactivate it, the 1K and 2K are left disconnected.
The circuit to the right of the 3K is already a Thevenin, where the voltage is 6V and the equivalent resistance is 9 Kohms. Converting to a Norton, we get Norton current of 0.667 mA and a resistance of 9 Kohms.
Now combine the two Nortons. The total curent will be -10mA + 0.667mA = -9.33 mA. The total resistance is infinite in parallel with 9K, which is simply 9K.