Hi,

I have a question about problem 34 in the worksheet.
http://www.allaboutcircuits.com/worksheets/thev.html

I start by removing the load resistor and replace it with a short(wire).

So the short circuit current is: 10/5000 = 2mA

Then i simply use current division to find out the voltage over the load.

(2200/3200 * 2mA) * 1000 = 1,375v over the 1K load resistor. However this is wrong.

What am i missing?

 

You have mixed Thevenin and Norton theory.

For the correct application of the Thevenin theorem you need to find the open (unloaded) ouput voltage, not the shorted output current.
Then you need to find the equivalent resistance as seen from the output port with all the sources (the 10V) nullified (shorted in your case, as you have a voltage source).
Finally you replace the circuit with a voltage source of a value equal to the the voltage calculated above, in series with a resistance of a value equal the to the resistance calculated above.

 

Hello,

Thank you for your response.
The open circuit voltage is still 10V is it not?

 

Hm, not quite. It's a voltage divider comprised from the 2.2k and 5k resistance. Can you guess again?

 

Over the 5k resistor:
\(\frac{5000}{7200}\cdot 10 = 6,94V\)

Over the 2.2k resisitor:
\(\frac{2200}{7200}\cdot 10 = 3,05V\)

 

Correct. Isn't it obvious that the output of the unloaded circuit is the voltage over the 2.2k resistor?

 

It wasn't, but it is now..