Use of superposition in Simple Circuits and Norton equivalent

In summary, the conversation discusses using superposition to solve for ix in a circuit where the resistor R3 represents a light bulb. It also talks about finding the Norton equivalent and power delivered by the independent current source, iS2. The person working on the problem makes some revisions and corrections, ultimately solving the problem and finding that the power delivered by the 10[mA] Cs is approximately 8[mW]. They also mention using current mesh method when solving for Isc.
  • #1
rock42
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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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  • #2
You'll need to leave any dependent sources in place. Independent sources can be suppressed leaving one active at a time.
 

Related to Use of superposition in Simple Circuits and Norton equivalent

1. How is superposition used in simple circuits?

Superposition is a technique used in circuit analysis that involves breaking down a complex circuit into smaller, simpler circuits and analyzing the individual components separately. This allows for a more systematic and efficient approach to solving complex circuits.

2. What is the principle behind superposition in circuit analysis?

The principle behind superposition is that in a linear circuit, the response at any point can be determined by adding the individual responses from each source acting alone. This is based on the linearity of the circuit components, where the behavior of each component is independent of the other components in the circuit.

3. What is the significance of Norton equivalent in circuit analysis?

Norton equivalent is a simplified representation of a complex circuit, where the current source and resistance are replaced by a single equivalent current source. This allows for easier analysis and calculation of circuit parameters, such as voltage and current, without having to deal with multiple sources and resistances.

4. How is Norton equivalent related to superposition in circuit analysis?

Norton equivalent is based on the principle of superposition, where the equivalent current source is determined by adding the individual currents from each source acting alone. This makes it a useful tool in applying the superposition technique to complex circuits.

5. Can superposition be used in non-linear circuits?

No, superposition can only be applied in linear circuits where the behavior of each component is independent of the other components. In non-linear circuits, the behavior of the components is dependent on the overall circuit, so the superposition principle cannot be used.

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