Thévenin equivalent

  • Thread starter gfd43tg
  • Start date
  • #1
gfd43tg
Gold Member
953
49

Homework Statement


Find the Thevenin equivalent circuit at terminals (a,b) of the circuit below:


Homework Equations





The Attempt at a Solution


my currents aren't adding up right. When I try to solve for Ix using my two equations that should find it, I get different answers.
 

Attachments

  • 3.10.png
    3.10.png
    16.4 KB · Views: 365
  • 3.10 attempt 1.pdf
    175.3 KB · Views: 122

Answers and Replies

  • #2
The Electrician
Gold Member
1,285
169
When you add the 1 volt test voltage at terminals a-b, you no longer have 4 meshes; you now have 5.

Your top schematic shows 4 meshes, and this allows you to calculate Vth, because the 2Ω resistor has no effect when the output is open circuited. Vth is just the voltage across the 8Ω resistor.

But as soon as you add the 1 volt test voltage as in the bottom schematic, you have 5 meshes. You need the current in that 5th mesh to calculate Rth.

You could solve it with only 4 meshes by assuming a short across a-b; this then replaces the 8Ω resistor with the parallel combination of 2Ω and 8Ω. Solving your 4 equations would give you the current through that parallel combination. Then use the current divider rule to calculate the current through the 2Ω resistor; that would be the short circuit current through a-b.

If you solve this using the nodal method, you only have 3 equations to deal with. That's what I meant when I said "You usually can use any network analysis method. One method may be easier for a given circuit than another."
 
  • #3
gfd43tg
Gold Member
953
49
I'm doing the proper amount of meshes, and now my voltage is high and the resistance seems way too high. I just put those equations into a 5x5 and 4x4 solver. Here is my 2nd attempt at it
 

Attachments

  • 3.10 attempt 2.pdf
    210.2 KB · Views: 130
  • #4
The Electrician
Gold Member
1,285
169
Your 4th equation is incorrect. Your units are wrong. I3 is a current and 8*I4 is a voltage.

What you need is to say I4-I3 = 2*Ix, then set Ix = I1-I2

Give that a try, and make the same correction in your 5 mesh equations.

Your 5th equation for the 5 mesh system should be -8*I4 + 10*I5 = -1
 
  • #5
gfd43tg
Gold Member
953
49
With the dependent current source there I don't know how to express the voltage drop across it. I redid it and got 75.4 V. my first 3 equations are unchanged but the last two are now

2 i1 - 2 i2 + i3 - i4 = 0

then for the 5 mesh system similarily, I changed to the same and what you said about -8 i4 + 10i5 = -1
 
Last edited:
  • #6
The Electrician
Gold Member
1,285
169
The voltage across the dependent current source is the same as the voltage across the 8Ω resistor in the 4 mesh case; that would be 8*I4 in the 4 mesh case. Your last two equations look good now, but I don't get 75.4 volts for that voltage.

Show the 4 equations for the 4 mesh case and the values you get for the 4 currents, and I should be able to see what the difficulty is.

Also, what do you get for the currents in the 5 mesh case?
 

Related Threads on Thévenin equivalent

Replies
12
Views
984
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
825
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
596
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
624
Top