Discussion in 'Feedback and Suggestions' started by Jony130, Jan 13, 2010.

Feb 17, 2009
3,993
1,116
2. ### Dcrunkilton E-book Co-ordinator

Jul 31, 2004
416
11
To determine the Thevenin equivalent voltage, reduce the source to zero. That is, replace the voltage source by a short circuit. The resistance between the outptu terminals is then (2K+400) || 1K || 300 = 210.53

3. ### kubeek AAC Fanatic!

Sep 20, 2005
4,689
806
No it´s not. When you short the supply, 300 becomes parallel to 1K, then this in series with 400 and finall parallel to 2K.
The answer is ((300||1k) + 400) || 2K, that is 480Ω.

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
OK,

300Ω !! 1KΩ = 230.769Ω

230.769Ω !! (2KΩ+400Ω) = 228.571Ω

Which is wrong. The fallicy is the 2KΩ is across the measuring point, and the 400Ω is part of the other network as seen by the measuring point. The resistance is being calculated is from the junction of 300Ω/400Ω/1KΩ.

Redrawing this schematic for clarity.

From the measurement point:

1KΩ !! 300Ω = 230.769Ω

(230.769Ω + 400Ω) !! 2KΩ = 479.532Ω

Resistance as seen from the power supply is

300Ω + (1KΩ !! (400Ω + 2KΩ)) = 1005.882Ω

15V Current is 15V / 1005.882Ω = 14.912ma

Drop across the 1KΩ is 10.526V

Drop across the 2KΩ is (10.526V 2KΩ) / (400Ω + 2KΩ) = 7.31V

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Last edited: Jan 19, 2010
5. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
3,993
1,116
The Eth voltage is OK.
Total resistance seen by voltage source (1KΩ) is is different than resistance see from the measurement point (480Ω).

6. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
OK, going through the math I did make a mistake.

I'm going to move this to my original thread.

Care to check my math?

Last edited: Jan 19, 2010
7. ### Dcrunkilton E-book Co-ordinator

Jul 31, 2004
416
11
I now agree with the proposed 479.53 Ohm Thevenin resistance.

I get 8.77 V Thevenin resistance as in the original problem answer.

Since I do not have access to the source files for the problems at ibiblio, I will not be able to do anything about changing the answer. It is part of Tony Kuphaldt's Socratic Electronics project.

8. ### ebot9000 New Member

Jan 1, 2010
1
0
Wow, this was just driving me nuts. Glad I'm not the only one. I'm getting Vth = 8.772 Volts / Rth = 479.53 Ω.

Look forward to clarification if the answer actually is Rth = 210.53 Ω, but I can't seem to get it.

When my answer didn't correspond with the posted answer, I did a simulation and put a test voltage across the measurement points, and the resulting current corresponded with a seen resistance of around 479.5 Ω.

But since I'm just learning this stuff, I'm ready to stand corrected!
-Ethan