Introductory circuit analysis - thevenin circuits

In summary: Now my 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 right1. 7.8kI1 - 5.6kI2 = 342. -5.6kI1 + 15.7kI2 = -18solving these equations will give I2 = 0.549mabthus Eth = 6.8k(0.549m) + 6 = 3.73 + 6 = 9.73v
  • #1
Mohdoo
20
0

Homework Statement




otkDF.jpg


Homework Equations



Thevenin stuff, right?

The Attempt at a Solution




So, I tried to re-write everything and make it a bit easier to understand:

yqoJT.jpg


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.

PNT40.jpg


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? =(
 
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  • #2
Mohdoo said:
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.Help? =(

You mean you are using the superposition principle?

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..
 
Last edited:
  • #3
rootX said:
You mean you are using the superposition principle?

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..

My apologies. I meant to label the two ends of "Rl" as A and B. I was told that using superposition, I am supposed to do that, and find Eth as the voltage between A and B.

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
Mohdoo said:
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?

This is what I used:
00498.png


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
rootX said:
This is what I used:
00498.png





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.

Ok, I think I got it!

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:
54Fwn.jpg


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 sort of right?

Once again, I really appreciate your help. You've been a huge help to me.
 
  • #6
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
Thanks for the replies everyone! I get it :D
 

Related to Introductory circuit analysis - thevenin circuits

1. What is a Thevenin circuit?

A Thevenin circuit is a simplified representation of a more complex circuit, where the circuit is reduced to a single voltage source and a single resistor. This allows for easier analysis and calculations.

2. How do I find the Thevenin equivalent voltage and resistance?

To find the Thevenin equivalent voltage, you can either use a voltage divider or measure the open-circuit voltage at the output terminals. To find the Thevenin equivalent resistance, you can either use a current divider or measure the resistance at the output terminals with all sources removed.

3. Can Thevenin circuits be used for non-linear circuits?

No, Thevenin circuits can only be used for linear circuits, where the components follow Ohm's Law. Non-linear components, such as diodes and transistors, cannot be represented by a single voltage source and resistor.

4. What is the purpose of Thevenin circuits?

The purpose of Thevenin circuits is to simplify complex circuits for analysis and design. They allow for easier calculations and can help determine the behavior of a circuit under different conditions.

5. Are there any limitations to using Thevenin circuits?

Thevenin circuits are only accurate for linear circuits and may not accurately represent the behavior of non-linear components. They also cannot account for any changes in the circuit due to varying conditions, such as changes in temperature or component values.

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