(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement

7. In the circuit below, the resistor R3 models a light bulb.

a) Use superposition to find iX.

b) Find the Norton equivalent as seen by the independent current source, iS2.

c) Find the power delivered by the independent current source, iS2, in the circuit below.

Diagram attached!

The attempt at a solution

I have yet to learn superposition in class, so forgive me if I am more than a little off the mark, but for part A I removed all sources but one, replacing the Vs with wires and the Cs with breaks.

12V left in: ix = 12/15000 [Amps]

5V left in: ix = 5/15000 [Amps]

.01A left in: ix = -.01 [Amps]

Dependent sources, when in a circuit by themselves, result in a zero current. Correct?

By Superposition ix= 17/15000 [amps] (Part A)

Removing the 10[mA] Cs, I now am searching for V(open circuit) and i(short Circuit).

I suppose this is wear my questions kick in (assuming I was correct before regarding superposition). Can I say that this new ix (it has changed since I removed the 10[mA] Cs) is the old ix plus .01[A]? Reworking superposition tells me I can, but I may be wrong entirely regarding superposition.

Regardless, Solving for Voc and Isc gives my Norton/Thev circuit that is incorrect. (I have selected answers and I am told that the power delivered by the 10[mA] Cs is ~8[mW])

Edit the First:

I have revised my answer for part A to 133/75000 [A] after discovering that dependent sources are to be left in when computing superposition.

Edit the Second:

I have checked my answer for A and found it to be correct via node-voltage method. For some reason I am still getting incorrect Thev/Norton circuits.

Edit the Third:

I have solved the problem, I was making an odd error solving for the short circuit current that was ultimately resolved by redoing the problem and using current mesh method when solving for Isc. The result came to be exactly 8[mW] not an approximation.

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# Use of superposition in Simple Circuits and Norton equivalent

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