Bogged down on Norton equivalent problem

In summary, the conversation discusses the effects of an 8ma ideal current source and a 20kohm resistor in parallel on the voltage drop across nodes A and B. The concept of Rth is also mentioned, as well as the role of the 30V voltage source as a barrier. The conclusion is that the current will stay on the left to satisfy KCL and the voltage source will not change its voltage.
  • #1
MotoPayton
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0
Picture an 8ma ideal current source pointing downward in parallel to a 20kohm resistor on the left of this picture. Wouldn't these elements contribute to the voltage drop across A and B.

Rth makes sense because the ideal voltage source will be shorted eliminating the last resistor but the voltage is confusing me.

Thanks for the help

Screenshot2011-10-04at50201PM.png
 
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  • #2
MotoPayton said:
Picture an 8ma ideal current source pointing downward in parallel to a 20kohm resistor on the left of this picture. Wouldn't these elements contribute to the voltage drop across A and B.

Rth makes sense because the ideal voltage source will be shorted eliminating the last resistor but the voltage is confusing me.

Thanks for the help

Screenshot2011-10-04at50201PM.png

You mean like this:

attachment.php?attachmentid=39635&stc=1&d=1317828530.gif


The 30V voltage source is not going to change its voltage no matter what. So any KVL equations you care to write concerning what's to the right of that source are going to have 30V there. Nothing (physically realizable) that happens to the left of the voltage source can affect that.

So the voltage source acts as an effective "barrier" to what goes on to the left hand and right hand sides of it.
 

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  • #3
Ya same picture except the 8mA is pointing downward..

Allright you so since it has to be 30V on the upper and lower nodes nothing can change that. And you can find the current though the 20kohm resistor to be 1.5mA.

Doing KCL on the lower node 6.5mA is moving to the right in the lower node. That 6.5 A will move through the voltage source since they are modeled as having 0 Resistance.

So all the current will stay on the left to satisfy KCL and there is no way 30V is going to change therefore you can throw the stuff on the left out.
 

FAQ: Bogged down on Norton equivalent problem

1. What is a Norton equivalent circuit?

A Norton equivalent circuit is a simplified representation of a more complex electrical circuit that contains a current source, a resistor, and a load. It is used to analyze and understand the behavior of a circuit without having to consider its internal components.

2. How is a Norton equivalent circuit different from a Thevenin equivalent circuit?

While a Norton equivalent circuit contains a current source and a resistor, a Thevenin equivalent circuit contains a voltage source and a resistor. The main difference is that the Norton equivalent circuit is used to analyze the behavior of a circuit in terms of current, while the Thevenin equivalent circuit is used to analyze the behavior in terms of voltage.

3. How do I find the Norton equivalent circuit of a complex circuit?

To find the Norton equivalent circuit, you will need to calculate the total current and the equivalent resistance of the circuit. The equivalent resistance is determined by short-circuiting all voltage sources and open-circuiting all current sources. The Norton current is equal to the total current, and the Norton resistance is equal to the equivalent resistance.

4. What is the purpose of using a Norton equivalent circuit?

The main purpose of using a Norton equivalent circuit is to simplify the analysis of a complex circuit. It allows us to focus on the behavior of the circuit in terms of current, which can be easier to understand and work with compared to analyzing the behavior in terms of voltage.

5. Can I convert a Norton equivalent circuit to a Thevenin equivalent circuit?

Yes, it is possible to convert a Norton equivalent circuit to a Thevenin equivalent circuit and vice versa. This can be done by using the conversion formulas, which involve converting the current source to a voltage source and the resistor to an equivalent resistor. However, the analysis and understanding of the circuit may differ depending on which equivalent circuit is used.

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