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# Finding Thévenin Equivalent Circuits

Module by: Don Johnson. E-mail the author

Summary: Introduction of Thévenin equivalent circuits.

For any circuit containing resistors and sources, the v-i relation will be of the form

v= R eq i+ v eq v R eq i v eq
(1)
and the Thévenin equivalent circuit for any such circuit is that of Figure 1. This equivalence applies no matter how many sources or resistors may be present in the circuit. In an example, we know the circuit's construction and element values, and derive the equivalent source and resistance. Because Thévenin's theorem applies in general, we should be able to make measurements or calculations only from the terminals to determine the equivalent circuit.

To be more specific, consider the equivalent circuit of Figure 1. Let the terminals be open-circuited, which has the effect of setting the current i i to zero. Because no current flows through the resistor, the voltage across it is zero (remember, Ohm's Law says that v=Ri v R i ). Consequently, by applying KVL we have that the so-called open-circuit voltage v oc v oc equals the Thévenin equivalent voltage. Now consider the situation when we set the terminal voltage to zero (short-circuit it) and measure the resulting current. Referring to the equivalent circuit, the source voltage now appears entirely across the resistor, leaving the short-circuit current to be i sq = v eq R eq i sq v eq R eq . From this property, we can determine the equivalent resistance.

v eq = v oc v eq v oc
(2)
R eq = v oc i sc R eq v oc i sc
(3)

## Exercise 1

Use the open/short-circuit approach to derive the Thévenin equivalent of the circuit shown in Figure 2.

### Solution

v oc = R 2 R 1 + R 2 v in v oc R 2 R 1 R 2 v in and i sc = v in R 1 i sc v in R 1 (resistor R 2 R 2 is shorted out in this case). Thus, v eq = R 2 R 1 + R 2 v in v eq R 2 R 1 R 2 v in and R eq = R 1 R 2 R 1 + R 2 R eq R 1 R 2 R 1 R 2 .

## Example 1

For the depicted circuit, let's derive its Thévenin equivalent two different ways. Starting with the open/short-circuit approach, let's first find the open-circuit voltage v oc v oc . We have a current divider relationship as R 1 R 1 is in parallel with the series combination of R 2 R 2 and R 3 R 3 . Thus, v oc = i in R 3 R 1 R 1 + R 2 + R 3 v oc i in R 3 R 1 R 1 R 2 R 3 . When we short-circuit the terminals, no voltage appears across R 3 R 3 , and thus no current flows through it. In short, R 3 R 3 does not affect the short-circuit current, and can be eliminated. We again have a current divider relationship: i sc = i in R 1 R 1 + R 2 i sc i in R 1 R 1 R 2 . Thus, the Thévenin equivalent resistance is R 3 ( R 1 + R 2 ) R 1 + R 2 + R 3 R 3 R 1 R 2 R 1 R 2 R 3 .

To verify, let's find the equivalent resistance by reaching inside the circuit and setting the current source to zero. Because the current is now zero, we can replace the current source by an open circuit. From the viewpoint of the terminals, resistor R 3 R 3 is now in parallel with the series combination of R 1 R 1 and R 2 R 2 . Thus, R eq = R 3 R 1 + R 2 R eq R 3 R 1 R 2 , and we obtain the same result.

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