Thevenin's and Norton's Theorem

Thevenin's and Norton's Theorem

Thevenin's Theorem

Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r. The value of e is the open circuit voltage at the terminals, and the value of r is e divided by the current with the terminals short circuited.

Sample Problem

Find the Thévenin equivalent with respect to the 7k ohm resistor.

Solution

Remove the 7k ohm, since it is not part of the circuit we wish to simplify. Keep the terminals open since we are finding the Thevenin.

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Find Vth, the voltage across the terminals (in this case it is the voltage over the 3k ohm). Combine the 1k and 2k in parallel.

1k || 2k = (1k*2k) / (1k+2k) = 2M/3k = 2/3k = 667 ohms

Now use a voltage divider to compute Vth across the 3k ohm.

Vth = [3k/(667+3k)] * 5V = 4.1V

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Find the Thevenin Resistance by deactivating all sources and computing the total resistance across the terminals. The voltage sources is shorted, as shown:

Now let's redraw the circuit, bringing the 1k and 2k into a vertical position (but still keeping them connected the same way electrically).

They are all in parallel, so:

Rth = 1k || 2k || 3k = 1 / (1/1k + 1/2k + 1/3k) = 545 ohms

Note, as a check, the equivalent resistance for parallel resistors is always smaller than the smallest resistor in the combination. For example, 545 is smaller than 1k.

The final Thevenin equivalent is then:

Norton's Theorem

Any collection of batteries and resistances with two terminals is electrically equivalent to an ideal current source i in parallel with a single resistor r. The value of r is the same as that in the Thevenin equivalent and the current i can be found by dividing the open circuit voltage by r.

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

Solution

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:

• mesh-current analysis
• superposition

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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: