Finding the magnitude and direction of current thro Kirchhoff law

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Homework Help Overview

The discussion revolves around applying Kirchhoff's laws to determine the magnitude and direction of current through a 10 ohm resistor. The original poster presents a discrepancy between their calculated current and the answer provided in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve the problem using their own method but questions where they might have gone wrong compared to the textbook answer. Other participants discuss the validity of the original poster's calculations and provide an alternative approach using the superposition principle.

Discussion Status

The discussion includes various perspectives on the calculations, with some participants affirming the original poster's work while others suggest that the textbook answer may be incorrect. There is an ongoing exploration of different methods to analyze the circuit.

Contextual Notes

The original poster references an attachment for their calculations, which may contain additional details relevant to the problem. There is also mention of specific voltage sources being shorted during the analysis, indicating a methodical approach to the problem.

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Homework Statement



Revered Members,
Please go through my attachment. I have been asked to find the magnitude and direction of current through the 10 ohm resistor. The answer that is given in the book is 1/23 Ampere from A to B.
But i solved as given in the attachment but i got the answer as 5/37 Ampere which is I3...the current which I assumed to pass through 10 ohm resistor. Since Please help me where i went wrong.

Homework Equations





The Attempt at a Solution


 

Attachments

  • prob.jpg
    prob.jpg
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Your work and results look fine. Must be an error in the book's answer.
 
I used the superposition principle as follows:

1. Short the 2V source & get voltage across the 10Ω = +2.1622V.
2. Short the 8V source and get same voltage = -0.8108V
3. Superpose the voltages to get +1.3514V across the 10Ω.
4. Divide by 10Ω to get i3 = 0.13514A which agrees with both of you!
 
Thank you gneill and rude man for your help.
 

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