Calculating Current in a Circuit with R1 = 10 Ohms and R2 = 8 Ohms

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Discussion Overview

The discussion revolves around calculating the current through resistor R1 in a circuit with given resistances R1 = 10 Ohms and R2 = 8 Ohms, a current source of 9A, and a voltage of 18V. The focus is on applying circuit analysis techniques, including Kirchhoff's laws and superposition.

Discussion Character

  • Homework-related, Technical explanation

Main Points Raised

  • One participant attempts to find the current through R1 by removing the current source, resulting in a calculated current of 2.25A.
  • Another participant suggests using Kirchhoff's laws, stating that the current entering any junction equals the current leaving, implying a method to find the current through R1.
  • A third participant mentions the importance of using superposition correctly, indicating that the current source should be treated as an open circuit rather than shorted, suggesting this may be a source of error in the initial approach.
  • A later reply acknowledges the oversight regarding the treatment of the current source, indicating a realization of the correct method.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the current through R1, with differing views on how to handle the current source and the application of circuit analysis techniques.

Contextual Notes

There are limitations regarding the assumptions made about the circuit configuration and the treatment of the current source, which remain unresolved in the discussion.

fan_boy17
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Homework Statement


the question is find the current through R1

R1= 10 Ohms R2=8 ohms
current source=9A
E=18V

Homework Equations


The Attempt at a Solution


I have tried removing the current source and i get a short circuit on R1 and hence i obtain a current of 2.25A
 

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What you are looking for are kirschoff's laws:
The current entering any junction is equal to the current leaving. So at the top I1 enters and use that together with the "voltage law" then you get the answer
 
When using superposition on current sources, you should open, instead of shorting, circuit that branch.

That may be where your problem lies.
 
Yes didn't realize that i needed to make the current source open circuit. thanks!
 

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