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### Network # Another Thevenin Question

M

#### Mrs. Kerchief

Jan 1, 1970
0
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.

K

#### Ken Taylor

Jan 1, 1970
0
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

J

#### JeffM

Jan 1, 1970
0
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).
Mrs. Kerchief

Thevenin Analysis
Rs
+----/\/\/\------------
|
-----
--- 15V
-----
---
|
=----------------------
Looking back with no load, we read 15V. So Vsource must be 15V.
(No current draw == no voltage drop across Rs == no voltage divider)

Rs
+----/\/\/\----+-------
| | ^
----- / |
--- \10k |
----- / 12V
--- \ |
| | v
+--------------+-------
12V / 10k = 1.2mA

(15V - 12V) / 1.2mA = 2k5
(what is dropped across Rs) - (current thru Rs) = Rs

M

#### Mrs. Kerchief

Jan 1, 1970
0
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.

(Mrs. Kerchief curtsies.) Mrs. Kerchief and tutor input 12 volts for
the output voltage. Thanks, Ken (and also Jeff).

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