Norton's Theorem

1. Jun 11, 2015

1. The problem statement, all variables and given/known data
Find Io using Norton's Theorem

2. Relevant equations

3. The attempt at a solution
I broke it into the following circuit:

I would like to break it down some more but if I open the node between 3k and 6k (ohms), and 6k and 6k (ohms), I run into problems....

What should I do to make sure I stick to Norton's Theor

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2. Jun 11, 2015

Staff: Mentor

I don't think you're restricted to using Norton's Theorem exclusively. It should be sufficient to determine the Norton equivalent of the network driving the current through the 6 k "load" using whatever means you wish. You might begin by suppressing all the sources and determining the Norton resistance.

By the way, 2 k + 3 k is not 6 k

3. Jun 11, 2015

FOIWATER

I just took a while and solved the problem by hand using mesh analysis with a supermesh and then put the circuit into QUCS and verified my answer.

I am having a hard time seeing how it would be easily done with norton's theorem or thevenin's theorem. I got a value for the equivalent resistance seen by the load resistor but breaking down the circuit to find the voltage across the terminals of the load resistor would lead me back to using mesh.

Last edited: Jun 11, 2015
4. Jun 11, 2015

Staff: Mentor

I'd go for a Thevenin approach and Nodal Analysis. Finding the Thevenin resistance (same as Norton resistance) is pretty easy with the sources suppressed. Removing the 6k load to open-circuit the output (For Thevenin you want the open-circuit voltage) leaves a single essential node. The rest is just finding particular potentials along branches.

5. Jun 11, 2015

FOIWATER

Yeah I agree that would be another way. converting the thevenin equivalent into a norton equivalent. The equivalent resistance is easy to find

The difference in voltages Vx and Vy is the thevenin equivalent voltage

6. Jun 12, 2015

At first I didn't know what you meant by "driving the current through the 6 k "load" using whatever means you wish." but with FOIWATER's picture:

I am able to understand now. Thank you! I will try my best to solve this using the techniques you suggested!

7. Jun 12, 2015

I was able to find the Voltage between the open circuit using Thevenin and Node analysis: I got Vth = ~10.8 V (making Vy facing positive and Vx near negative), and Rth = 33/5. So the short circuit current is 10.8/6.6=1.6464 and current division between the load and 6.6 KOhms makes the load current 0.857. This hsould be negative because of the way the current is facing and the way I had my voltages set up. Thanks everyone!

(Btw FOIWATER, your R6 resister should be 3k)

8. Jun 12, 2015

FOIWATER

Glad you figured it out friend.

9. Jun 12, 2015

Follow up question on using QUCS:
I'm trying to make a current controlled voltage source matching this diagram...:

Im not sure how exactly the current controlled voltage source is used in qucs.. Examples help, but I didnt know the right name for it in the examples given in qucs

Code (Text):

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<R R1 1 230 -450 15 -26 0 1 "3000 Ohm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
<R R2 1 440 -450 15 -26 0 1 "6000 Ohm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
<R R4 1 530 -630 -26 15 0 0 "2000 Ohm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
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<IProbe Pr1 1 600 -430 16 -26 1 3>
<GND * 1 430 -310 0 0 0 0>
<CCVS SRC1 1 340 -620 34 -26 1 3 "4000" 1 "4000" 0>
<.DC DC1 1 40 -540 0 43 0 0 "26.85" 0 "0.001" 0 "1 pA" 0 "1 uV" 0 "no" 0 "150" 0 "no" 0 "none" 0 "CroutLU" 0>
<Idc I2 1 420 -780 -26 18 0 0 "2 mA" 1>
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10. Jun 12, 2015

Staff: Mentor

R4 is shorted by your wiring of the controlled current source "sense line". Note that there is a path from the top of R5 to the top of R2 that does not pass through R4!

11. Jun 14, 2015