How Do You Simplify a Circuit Using Thevenin's Theorem?

[nobodyuknow]

Homework Statement

Above given is the circuit I am supposed to simplify where the 1/5Ω resistor is the load. I'm unsure how to convert this circuit into an equivalent Thevenin Circuit.

Homework Equations

V = I*R (Ohm's Law)

The Attempt at a Solution

Using Microcap and it's Dynamic DC Analysis Tool, I measured the current flow at the load. I managed to break it down to this:

http://img844.imageshack.us/img844/282/5x8t.png

Where V5 was a 2/3V voltage source. (2A current source in series with the 1/3Ω resistor).

If this is correct, I would then say that the 2/3V voltage source and the 4A current source were in series and that the V4 voltage source and the 1/6Ω resistor were in series. Is this the correct approach?

 

[gneill]

Remember that a Thevenin equivalent is a single voltage source with a single series resistance (for DC circuits).

What's the usual algorithm for finding the Thevenin equivalent analytically?

 

[nobodyuknow]

Would the first step be setting all those current sources to 0? Thus left with the remaining voltage source and the two resistors?

http://img820.imageshack.us/img820/2318/q2z4.png

 

[gneill]

No. You must suppress ALL sources to find the Thevenin resistance: Remove all current sources and short all voltage sources.

To find the Thevenin voltage you must find the open-circuit potential across the load (remove the load and find the potential across the open terminals).

Your course notes or text should cover this!

 

[rezeew]

Yes, your approach is correct. To find the Thevenin equivalent circuit, you can follow these steps:

1. Identify the load resistor (1/5Ω in this case) and remove it from the circuit.

2. Calculate the open-circuit voltage (Voc) across the terminals where the load resistor was connected. This can be done by using the voltage divider rule, considering the 1/3Ω and 2/3V sources in series.

3. Calculate the short-circuit current (Isc) flowing through the terminals where the load resistor was connected. This can be done by using the current divider rule, considering the 1/6Ω and 4A sources in parallel.

4. The Thevenin equivalent voltage (Vth) will be equal to the open-circuit voltage (Voc) and the Thevenin equivalent resistance (Rth) will be equal to the ratio of the open-circuit voltage (Voc) to the short-circuit current (Isc).

5. Draw the Thevenin equivalent circuit with the Thevenin equivalent voltage (Vth) in series with the Thevenin equivalent resistance (Rth).

In this case, your Thevenin equivalent circuit would consist of a 2/3V voltage source in series with a 1/2Ω resistor. This is because the 2/3V voltage source and 4A current source are in series, and the 1/6Ω resistor and V4 voltage source are in parallel.

Hope this helps!