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Introductory circuit analysis  thevenin circuits 
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#1
Dec809, 12:36 AM

P: 20

1. The problem statement, all variables and given/known data
2. Relevant equations Thevenin stuff, right? 3. The attempt at a solution So, I tried to rewrite everything and make it a bit easier to understand: I ended up getting Rth completely fine and everything, but I can't seem to get Eth. So from there, I figured that I simply do like other problems that have 2 voltage sources. Simply act as if all the other sources are pieces of wire while leaving one at a time actually there, getting the voltage, etc. The only thing about this is, I assumed that the voltage of AB is the same as the voltage of the6.8K resistor. When I looked over other examples, all of which only had 2 voltage sources, the voltage of AB was simply the voltage of a resistor in the same place that the 6.8K resistor was. But the thing is, in those examples, there were only 2...Not sure if that messes things up =( So then because E1 had a positive polarity, E2 had a negative, and E3 had a positive, I added E1 and E2, then subtracted E2. When I did this, I simply got the completely wrong answer. Rth was fine, but Eth is just plain completely wrong. The book says that Eth should be 9.74V. I got like 12 something =( Help? =( 


#2
Dec809, 12:45 AM

P: 1,294

I am not sure what is AB .. you haven't labeled it anywhere. However, one simpler approach I can suggest is to convert each voltage and resistor line to a current source and resistor in parallel. Add up all the currents and add up all the resistors also in ll.. (sorry, missed the 3.3kohm, can do this to the first two wires and then add 3.3.. and do the last wire). I was able to come with 9.76 number (few rounding problems) under 30 seconds through this approach however your approach requires some paper work.. 


#3
Dec809, 01:16 AM

P: 20

So what you did was simply make the first two wires current with resistors in ll? I'll give it a shot! I don't suppose you could post some of the formulas you used? Thanks again!! 


#4
Dec809, 01:29 AM

P: 1,294

Introductory circuit analysis  thevenin circuits
Your assumption is wrong that E_thevinan is equivalent to the voltage across 6.8 kOhms, it is equal tp the voltage across 6.8 kOhms + 6 V. 


#5
Dec809, 02:58 AM

P: 20

I think that my first 2 voltages were right. That is to say, for E1, and E2. I think my problem came from E3. If I am incorrect, please correct me. So what I changed was, well, I'll simply attach a picture of how I changed E3 and then added all the voltages together for Eth: The only thing is that I got about 9.84...Not 9.74. Did I just get lucky by getting close? Or am I almost sorta right? Once again, I really appreciate your help. You've been a huge help to me. 


#6
Dec2009, 09:13 AM

P: 4

Eth is measured from voltage drop of the 6.8kohm resistor and the 6v source so discard the 1.2kohm resistor and RL
now your circuit has only two meshes instead of three so by kvl the equations below were derived. I1 mesh current on the left and I2 mesh current on the right 1. 7.8kI1  5.6kI2 = 34 2. 5.6kI1 + 15.7kI2 = 18 solving these equations will give I2 = 0.549mA thus Eth = 6.8k(0.549m) + 6 = 3.73 + 6 = 9.73v 


#7
Dec2009, 07:17 PM

P: 20

Thanks for the replies everyone! I get it :D



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