PHYS345 Electricity and Electronics

Detailed Thevenin Generator Example

We have been considering the following multiloop circuit. Let's determine the current flowing through the 6k resistor using the Thevenin approach. Determining the Thevenin 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 Thevenin Approach

Redrawn to emphasize Thevenin approach

Thevenin 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 Thevenin Resistance

Thus,

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

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

Sources killed for Thevenin resistance determination

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

Thevenin Generator Ready for Use!

Thevenin generator ready for use

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

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

Usefulness of the Thevenin Approach?

So when is the Thevenin 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 Thevenin 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 Thevenin approach will be conceptual, as we replace two-terminal devices that we encounter with much simpler Thevenin 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.