# Norton's Theorem - RLC

Discussion in 'Homework Help' started by smarch, Jun 23, 2010.

1. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
Hi,

I am having trouble with this question:

Using Nortons theorem, determine the current in the 15Ω resistor of the AC circuit.

I have attached the schematic in the post.

Do I take the norton equivalent of the left hand side of the circuit, then add the currents of the current sources together?

Any help would be appreciated.

thanks.

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2. ### Accipiter New Member

Sep 20, 2009
9
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Yes, it just requires a few steps. You basically just need to flip-flop between voltage and current source equivalents a couple of times (thevenin and norton representations). Add current sources in parallel. Try it out...

3. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0

A couple of times?

Can you not convert the left hand side to current source and resistor in parallel and then add the current sources?
Or is there more to it?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Have you considered using superposition?

5. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
No, the questions says to use Nortons theorem. Thanks though.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Yes - that's the approach using Norton's theorem. Both the Norton equivalent current source and parallel impedance values will be complex. So adding the two current sources will involve complex addition. Also requires complex number manipulation to find the current split into the 15Ω.

7. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
I have reduced the circuit down and I have got two parallel current sources with a parallel complex resistor in between them.
I can't just add the current sources with this resistor in between them can I?
So do I transform the first current source to a voltage source in series with the resistor and find that current through the resistor and add it to the second current source?

8. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
If two current sources are parallel (touch the same nodes), then you can add them. In the step before the end, you should have a current source with an impedance parallel to it, and parallel to all that a series of the 15Ohm coil and the resistance in question. Convert the current source to its Thevenin equivalent, and now you should have a voltage source and 3 elements in series with it. It will be all clear by then.