# thevenin resistance question

Discussion in 'Homework Help' started by phenohm, Nov 25, 2006.

1. ### phenohm Thread Starter Member

Nov 12, 2006
10
0
hello all! thanks for your help in the past everybody, i'm looking forward to another bailout.

i'm a little confused about how to calculate thevenin resistance... my only material is the "lessons in electronic circuits" books, and they're a little light in that spot i think, at least for me. i've included pictures of two questions, could someone write out the steps to find the resistance? i'm getting the voltage correct, but while i've been getting the resistance correct in simpler networks i think i have been doing it in kind of a hacky way - finding the amperage through that spot, short circuit style, and figuring out what the resistance is from there. that's not working for me in a more complex network, as in q2, and doesn't make sense in a current-source network, as in q1.

thanks for the help!

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2. ### kc8ljh Member

Nov 25, 2006
15
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1) remove the load from the circuit
2) label the 2 terminals a and b etc,,
3) set sources to zero, since we have a current source we replace it with an open
4) determine Rth we have both resistors in series so Rth = 8k + 3.3K +11.3K
5) replace sources that were removed in step 3 and do Vth, the 3mA goes through the 11.3kohms so V = I*R = 3mA* 11.3k = 33.9v for both resistors. and looking at the schematic that would be Vth = 33.9V

3. ### phenohm Thread Starter Member

Nov 12, 2006
10
0
thanks k, that helps a lot with the first problem. i didn't know to replace a current source with a jumper. any ideas for the second problem? i just sat down with it again and my best guess for the resistance was wrong:

((1k//300)--400)//2k = 479.53ohms

and when i calculate the open circuit and closed circuit numbers and work out the big e-i-r number tables, i get 480ohms. so, at least i'm consistent. the answers tell me i should get 200-some ohms. what am i doing wrong?

4. ### kc8ljh Member

Nov 25, 2006
15
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I got that too for 2nd schematic. the 300 and 1k are in parallel, and the 400 is in series with the 300 and 1k, and the 2k is in parallel with them. And for Vth i got 8.75v

5. ### phenohm Thread Starter Member

Nov 12, 2006
10
0
thanks so much for working that out for me k, 8.77 is what i got for Vth as well. i guess the answer is wrong, maybe i'll email whoever is in charge of the socratic electronics project and see if they know something we don't about the problem, or if there's an error.

6. ### farmosh203 Member

Nov 26, 2006
20
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For your second question, I just wanted to show you a way that you can check your answer. Sometimes I get a little confused on what is in parallel and series, so I found that this method works. At the breaking point of the circuit, make a voltage source with any arbitrary value "Voltage test". Usually I just use 1 Volt. Then short circuit all voltage sources and open circuit all current sources. Find the current through the Test Voltage source and use V = IR to find R_thevenin.

I did an example of the work on how you would do this in the attachment.

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7. ### phenohm Thread Starter Member

Nov 12, 2006
10
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awesome. i will *definitely* use that, thanks for writing it out once for me.