What is the Norton equivalent current for terminals a and b in this circuit?

In summary, when finding the norton equivalent with respect to terminals a and b, the current i4 will be equal to the current flowing through the short circuit between a and b. Additionally, when shorting a and b, the voltage across the 30k ohm resistor will be 0.
  • #1
gomezfx
20
0

Homework Statement


Find the norton equivalent with respect to terminals a and b
[PLAIN]http://img821.imageshack.us/img821/7177/nortonequivalent.jpg

Homework Equations


KVL, KCL

The Attempt at a Solution


I set i1 = .008 A
Then I found the rest of the currents.

KVL@ loop i2:
2000(i2-i1)+30=0
i2 = .0065 A

KVL@ loop i3:
-30+15000i3=0
i3=.002 A

KCL@ node D
i3-i4-.010=0
(.002)-i4-.010=0
i4= -.008 A

KVL@ loop iab:
30000(iab-i4)=0
iab= -.008A

I got the same current for i4 and iab, unless I did my math wrong.
Can someone check my equations and set up to make sure I'm even finding the right current for the norton equivalent circuit?

EDIT. I redid i4 by doing (i4-i3)=.010 A
I got i4 = .012 A so I'm not sure which current to o use.
 
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  • #2
There's nothing wrong with having iab = i4. The current will choose the path of least resistance, so if you short a and b, all the current flows through the short. Equivalently, you can think of it this way. If you short a and b, the voltage across the 30k-ohm resistor will be 0, so no current flows through it. Applying KCL to E then gives you that iab = i4.

Your initial attempt to find i4 is correct. In the second try, you introduced a sign error. At node D, i3 enters and i4 and 10 mA exit, so you have i3 = i4 + 0.010 A.

I didn't check everything in the rest of the problem, but I'll note that given how you drew the direction of i1, you should have used i1=-0.008 A.
 
  • #3
How come I can just assume that the 30k ohm resistor will be 0 when I short a and b. My professor made did this same thing in class with another problem and a lot of things I've been reading do the same.

I'll redo the problem with the -0.008 A as well.
 
  • #4
gomezfx said:
How come I can just assume that the 30k ohm resistor will be 0 when I short a and b.
It doesn't make sense when you say the "resistor will be 0." What is that supposed to mean?
 
  • #5
vela said:
It doesn't make sense when you say the "resistor will be 0." What is that supposed to mean?

Sorry, I meant to say that the voltage across the resistor will be 0.
 
  • #6
Oh, OK. When you short a and b, the two ends of the resistor are connected to the same node in the circuit, so the voltage across the resistor must be 0.
 

Related to What is the Norton equivalent current for terminals a and b in this circuit?

1. What is a Norton Equivalent Circuit?

A Norton Equivalent Circuit is a simplification of a more complex electrical circuit that contains a current source and a parallel resistance. It is used to model the behavior of a circuit, making it easier to analyze and design.

2. How is the Norton Equivalent Circuit different from Thevenin Equivalent Circuit?

The main difference between the Norton and Thevenin Equivalent Circuits is the type of equivalent source used. The Norton Equivalent Circuit uses a current source, while Thevenin Equivalent Circuit uses a voltage source.

3. What is the purpose of finding a Norton Equivalent Circuit?

The main purpose of finding a Norton Equivalent Circuit is to simplify a complex circuit into a single current source and parallel resistance, making it easier to analyze and design. It also allows for the calculation of the maximum power transfer from the circuit to a load.

4. How do you calculate the Norton Equivalent Current and Resistance?

To calculate the Norton Equivalent Current, you can use the formula IN = ISC = VOC / REQ, where ISC is the short-circuit current, VOC is the open-circuit voltage, and REQ is the equivalent resistance. The equivalent resistance can be found by removing all sources from the circuit and calculating the total resistance between the two terminals.

5. How does the Norton Equivalent Circuit affect the overall behavior of a circuit?

The Norton Equivalent Circuit does not affect the overall behavior of a circuit, as it is simply a simplification of the original circuit. It will accurately represent the behavior of the original circuit in terms of current and resistance. However, it may not accurately represent other properties of the circuit such as voltage or power.

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