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Hi! This is my first post so please go easy!

Im doing some circuit revision and came across this question that im having problems with. It says there are 3 possible values for the RL i)1ohm ii)4 ohm and iii)8 ohm: derive a norton equivalent and determine which of the loads develops the most power.

I have so far got:

But I think my method may be seriously flawed! I then used I=Vs/Rs+RL to get the different currents then P=I^2 R to get the Power dissapation. Any help and feedback would greatly appreciated!

Noob signing out! :D

Are you absolutely sure you have the circuit diagram of the original circuit correct?

If it is correct then you can disregard R1 since it has no effect in parallel with an ideal voltage source (i.e. a voltage source with zero source resistance ... think about it ... run some sims).

And why have you placed a short circuit across RL?

Also:

https://www.circuitlab.com/docs/the-basics/#ground

:)

These might help:

http://en.wikipedia.org/wiki/Norton's_theorem

http://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html

http://people.clarkson.edu/~jsvoboda/eta/dcWorkout/thevenin.pdf

and:

https://www.circuitlab.com/circuit/8teegk/thevenin-and-norton-equivalent-circuits-01/

Thanks for the quick reply! I realised that i really need much more revision! the diagram was correct, but the second is woefully wrong! I didnt short the resistor, like you said and not sure why i put the the short on the opposite side! I will post a link to my much revised and hopefully correct Norton Equiv circuit. Thanks again, much appreciated.

P.S I liked the links, saved to my bookmarks!

Next thing to do is add a ground:

https://www.circuitlab.com/docs/the-basics/#ground

Then run a DC sim of original and simplified circuits to check:

i) open circuit voltage V(a)-V(b);

ii) short circuit current (current through a short circuit applied across a and b, i.e. connect a 0V source across a and b and measure the current through it or set RL = 0 and measure the current through that: I(RL.nA) );

both (i) and (ii) should give the same result for both circuits.

You can do the short / open cct test in one DC Sweep sim by entering RL.R into the Parameter box and then entering a Custom list of resistor values.

:)