PHYS345 Electricity and Electronics

Detailed Thévenin Generator Example

We have been considering the following multiloop circuit. Let's determine the current flowing through the 6k resistor using the Thévenin approach. Determining the Thévenin generator that represents the circuit will facilitate our consideration of different values of resistance in place of the 6k.

The multiloop circuit we have been considering

Circuit Redrawn in Preparation for Thévenin Approach

Redrawn to emphasize Thevenin approach

Thévenin Generator

Thevenin generator

Determination of Open-Circuit Voltage

Configured for determination of open-circuit voltage

Applying nodal anaylsis:

(Voc-9)/5 + (Voc+9)/12 = 0
12 (Voc-9) + 5 (Voc+9) = 0
17 Voc = 63
VTh = Voc =3.71 V

Determination of Short-Circuit Current

Configured for determination of short-circuit current

Applying nodal anaylsis again:

-9/5k + 9/12k + Isc = 0
Isc = 1.05 mA

Determination of Thévenin Resistance

Thus,

RTh = Voc / Isc
RTh = (3.71 V) / (1.05 mA)
RTh = 3.53 k

Alternatively, killing the sources for determination of the Thévenin resistance:

Sources killed for Thevenin resistance determination

5k in parallel with 12k yields an equivalent resistance of 3.53k!

Thévenin Generator Ready for Use!

Thevenin generator ready for use

Now the current can be found by adding the load resistance and the Thévenin resistance to find the equivalent resistance for the circuit. Application of Ohm's law using the equivalent resistance and the Thévenin voltage then yields the current.

I = (3.71 V) / (3.53k+6.00k)
I = 0.39 mA.

Usefulness of the Thévenin Approach?

So when is the Thévenin approach useful? One application would be for evaulating the effects of tolerance on resistor specifications. Let's say that the 6k resistor is specified as having a 10% tolerance. That is, a resistor marked as 6k and placed in the circuit could actually have a resistance between 5.4k and 6.6k. Using the circuit analysis approaches that we studied earlier might necessitate lenghty recalculation for the additional two values as we set out to determine how the current will be effected by this variation. With the Thévenin approach, we need only reconsider the two equivalent resistances, 8.93k and 10.1k, and the subsequent currents, 0.366 mA and 0.415 mA, to find that a 10% variation in resistance will yield about a 6% variation in current.

The other main use of the Thévenin approach will be conceptual, as we replace two-terminal devices that we encounter with much simpler Thévenin generators. Stay tuned....


"http://www.physics.udel.edu/~watson/phys345/examples/thevenin-example.html"
Last updated Sept. 15, 1998.
Copyright George Watson, Univ. of Delaware, 1998.