Mrs. Kerchief said:

There's a self-test at the end of each chapter of my EE textbook, and

I couldn't figure out--because I didn't understand--how the correct

response was arrived at in the following question.

"A certain circuit produces 15 V across its open output terminals, and

when a 10K load is connected across its output terminals, it produces

12 V. The Thevenin equivalent for this circuit is...(and the answer

is 15 V, 2.5K Ohms).

I didn't know where to begin to figure the answer out, and neither

does my tutor. My instructor goes way too fast for people taking

their first electronics course.

Anyway, thanks for any help.

In an ideal world a voltage source has no series resistance, so no matter

what load you put on, the voltage at the output is the same. The Thevenin

circuit assumes a simple series resistance, which then acts as a part of a

voltage divider at the output. Therefor if you put a 10k load on your 15V

source and see 12V at the output, you just transpose a voltage divider

equation to find your unknown resistance.

ie. Vo = Vs x (Rl/(Rl+Rth))

where Vo is the Output Voltage, Vs is the Source voltage, Rl is the Load

Resistance and Rth is the Thevenin equivalent series resistance.

Okey dokey?

Cheers.

Ken