Solving a Superposition Circuit Question: Finding Ia without a Voltage Source

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SUMMARY

The discussion centers on solving for the current Ia in a superposition circuit where the voltage source is shorted out. Participants confirm that it is indeed possible to find Ia using the superposition theorem, as well as Kirchhoff's and Ohm's laws. The equations derived include ib = -V1/3, ic = V2/2, and ia = (V1 - V2)/5, leading to a system of five equations with five unknowns. This method allows for a clear resolution of the circuit's parameters.

PREREQUISITES
  • Understanding of superposition theorem in circuit analysis
  • Familiarity with Kirchhoff's laws
  • Knowledge of Ohm's law
  • Ability to solve systems of linear equations
NEXT STEPS
  • Study the application of the superposition theorem in circuit analysis
  • Learn how to apply Kirchhoff's Voltage and Current Laws in complex circuits
  • Practice solving systems of equations using matrix methods
  • Explore advanced circuit analysis techniques, such as Thevenin's and Norton's theorems
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in circuit analysis and troubleshooting will benefit from this discussion.

andyintranzit
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http://img442.imageshack.us/img442/2070/circuit1vf1.gif

i drew this from a superposition question.. the voltage source is shorted out.

what on Earth could Ia be if there's no voltage source?
 
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wow I am really stupid. you have to solve for only one voltage/current source at a time, right?

just out of interest though... is it possible to solve that?
 
andyintranzit said:
wow I am really stupid. you have to solve for only one voltage/current source at a time, right?

just out of interest though... is it possible to solve that?
Of course you can solve it by using superposition. Otherwise, you can solve it using Kirchoff's and Ohm's laws.
Call V1 the voltage between resistors 3 and 5 ohm and V2 the voltage between 5 and 2 ohm. Call ib the current in the 3 ohm resistor and ic in the 2 ohm. You have:
ib = -V1/3
ic = V2/2
ia = (V1 - V2)/5
ib = ia + 6
ic = ia + 8
You have 5 independent equations in 5 unknowns. You can solve it for the unknowns.
 

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