# Worksheet thevenin help

Discussion in 'Homework Help' started by regexp, Dec 5, 2010.

1. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
Hi,

I have a question about problem 34 in the worksheet.

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?

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
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.

3. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
Hello,

The open circuit voltage is still 10V is it not?

4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Hm, not quite. It's a voltage divider comprised from the 2.2k and 5k resistance. Can you guess again?

5. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
Over the 5k resistor:
$\frac{5000}{7200}\cdot 10 = 6,94V$

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

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Correct. Isn't it obvious that the output of the unloaded circuit is the voltage over the 2.2k resistor?

7. ### regexp Thread Starter New Member

Nov 20, 2010
24
0
It wasn't, but it is now..