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.

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

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:

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