[Circuits] Calculating the Norton Equivalent

In summary: Ah, that makes sense. Yeah, I think I'm going to adopt the same convention to avoid these types of mistakes again.
  • #1
ainster31
158
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.
 
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  • #2
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
phinds said:
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
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
gneill said:
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
ainster31 said:
Is there something wrong with my first attempt?

Yes. Where did the 240V come from ?
 
  • #7
gneill said:
Yes. Where did the 240V come from ?

I typed it wrong. I have fixed it now. Sorry about that.
 
  • #8
ainster31 said:
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##
 
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  • #9
gneill said:
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.
 

1. What is the Norton equivalent circuit?

The Norton equivalent circuit is a simplified representation of a complex circuit that consists of a current source in parallel with a resistor. It is used to simplify circuit analysis and calculations by reducing the number of components needed to represent the original circuit.

2. How is the Norton equivalent current calculated?

The Norton equivalent current is calculated by taking the short-circuit current of the original circuit, which is the current that flows when the output terminals are connected together, and dividing it by the total resistance of the original circuit.

3. What is the purpose of calculating the Norton equivalent?

The purpose of calculating the Norton equivalent is to simplify complex circuits and make them easier to analyze. The equivalent circuit can be used to determine the behavior of the original circuit, such as voltage and current values, without having to work through the entire circuit calculations.

4. How is the Norton equivalent resistance determined?

The Norton equivalent resistance is determined by removing all voltage sources from the original circuit and calculating the total resistance between the output terminals. This can be done by using series and parallel resistor rules.

5. What are the limitations of the Norton equivalent circuit?

The Norton equivalent circuit is a simplified representation of a circuit and therefore it does not account for all the characteristics of the original circuit. It does not take into account reactive components, such as capacitors and inductors, and it is only valid for linear circuits. It also assumes that the original circuit is in steady state, meaning that the components are not changing over time.

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