[Circuits] Calculating the Norton Equivalent

  • Engineering
  • Thread starter ainster31
  • Start date
  • #1
157
1

Homework Statement



h_1390343963_6347373_a2b980a606.png


Homework Equations





The Attempt at a Solution



I am trying to compute ##I_N##.

I connected nodes a and b. The circuit after some simplification:

h_1390344359_4911840_0dbd81aae2.png


$$\frac { 180-{ V }_{ 1 } }{ 20 } =\frac { 120-V_{ 1 } }{ 40 } \\ { V }_{ 1 }=240\quad V\\ \\ R=\frac { V }{ I } \\ 40=\frac { 120 }{ I_{ N } } \\ I_{ N }=3\quad A$$

The correct answer is 1A.
 
Last edited:

Answers and Replies

  • #2
phinds
Science Advisor
Insights Author
Gold Member
17,057
8,075
Why would you replace the 3amp current source with a 120 volt voltage source (I KNOW why you did it, but it's wrong and I want you to think about it)
 
  • #3
157
1
Why would you replace the 3amp current source with a 120 volt voltage source (I KNOW why you did it, but it's wrong and I want you to think about it)

I did it to simplify the circuit and make the nodal analysis easier.
 
  • #4
gneill
Mentor
20,925
2,867
Actually, changing the 3A / 40Ω subcircuit to its Thevenin equivalent works here. You're left with a simple series circuit, so no need for nodal analysis. Just calculate the current from the given resistances and voltages.
 
  • #5
157
1
Actually, changing the 3A / 40Ω subcircuit to its Thevenin equivalent works here. You're left with a simple series circuit, so no need for nodal analysis. Just calculate the current from the given resistances and voltages.

Is there something wrong with my first attempt?
 
  • #6
gneill
Mentor
20,925
2,867
Is there something wrong with my first attempt?

Yes. Where did the 240V come from ?
 
  • #7
157
1
Yes. Where did the 240V come from ?

I typed it wrong. I have fixed it now. Sorry about that.
 
  • #8
gneill
Mentor
20,925
2,867
I typed it wrong. I have fixed it now. Sorry about that.

Okay, now you need to correct the signs in your node equation. The left hand side yields a current flowing into the node from the 180V supply. On the right hand side you should therefore have a current flowing from the node to the 120V supply. You've got the right hand side yielding a current flowing from the 120V supply to the V1 node instead.

I find it easier, and less prone to error, to write all node equations from the perspective of the node itself and assuming that all currents are flowing out of the node (my choice of direction; you could choose the opposite if you wish). Sum all these currents on the same side of the equals and equate to zero:

##\frac{V1 - 180}{20} + \frac{V1 - 120}{40} = 0##
 
  • #9
157
1
Okay, now you need to correct the signs in your node equation. The left hand side yields a current flowing into the node from the 180V supply. On the right hand side you should therefore have a current flowing from the node to the 120V supply. You've got the right hand side yielding a current flowing from the 120V supply to the V1 node instead.

I find it easier, and less prone to error, to write all node equations from the perspective of the node itself and assuming that all currents are flowing out of the node (my choice of direction; you could choose the opposite if you wish). Sum all these currents on the same side of the equals and equate to zero:

##\frac{V1 - 180}{20} + \frac{V1 - 120}{40} = 0##

Ah, that makes sense. Yeah, I think I'm going to adopt the same convention to avoid these types of mistakes again.
 

Related Threads on [Circuits] Calculating the Norton Equivalent

Replies
6
Views
4K
Replies
1
Views
540
Replies
11
Views
2K
Replies
8
Views
4K
  • Last Post
Replies
3
Views
1K
Replies
3
Views
2K
Top