# Norton's Theorem

Discussion in 'Homework Help' started by carelesswhisper, Apr 27, 2009.

1. ### carelesswhisper Thread Starter New Member

Apr 27, 2009
2
0
Hi:

I try to apply Norton"s theorem at a very simple circuit. The circuit contains a DC voltage source with two in parallel connected resistors, and this parallel combination is connected in series with the voltage source. I want to find the current throw one of these resistors. The short circuit current (In) is infinity????

How to continue??

Carelesswhisper

2. ### StayatHomeElectronics AAC Fanatic!

Sep 25, 2008
1,020
71
Can you put a copy of your circuit here for us to look at? Label the resistors and the voltage sources if you can. Then list what you did to get your answer.

When you are finding Thevenin or Norton Equivalent resistance you need to remove the load, set all independent voltage and current sources to zero, i.e. short voltage supplies and open current supplies, and then compute the resistance as seem by the load terminals.

To get an infinite current, one of the supplies is probably still present.

3. ### carelesswhisper Thread Starter New Member

Apr 27, 2009
2
0
here is my circuit

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4. ### StayatHomeElectronics AAC Fanatic!

Sep 25, 2008
1,020
71
R1 is simply another load on the power supply E and does not really play a role in the overall circuit equivalence. That you show by R(norton) = 0. You are, therefore, trying to get the Norton equivalent circuit for an ideal power supply, which I don't know that that is technically useful. I think that sort of by definition the equivalent circuit needs to have both I(norton) and R(norton) to be usefull.

If you are trying to get current in either resistance, then i = E / R works without any circuit equivalence needed. Manipulate the circuit drawing if you need to to make it clear.