How Much Power Is Dissipated in Each Resistor in a Multiloop Circuit?

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Homework Help Overview

The discussion revolves around a multiloop circuit containing three resistors (R1, R2, and R3) and focuses on determining the power dissipated in each resistor. The original poster raises questions about the application of loop and junction laws in circuit analysis.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Kirchhoff's laws to analyze the circuit but questions the directionality of the loops used in their equations. Some participants clarify that the direction of traversal does not need to be consistent across loops, as long as potential changes are correctly accounted for.

Discussion Status

Participants are actively engaging in clarifying the application of circuit laws. Guidance has been provided regarding the flexibility of loop direction, and suggestions have been made to simplify the analysis by using a different loop. The discussion appears to be productive, with participants exploring different interpretations of the problem.

Contextual Notes

The original poster's inquiry reflects a common point of confusion regarding the application of circuit analysis techniques, particularly in the context of homework constraints that may limit the resources available for assistance.

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Homework Statement


26-p-063-alt.gif


Three resistors, R1 = 32 Ω, R2 = 45 Ω, and R3 = 60 Ω, are connected in a multiloop circuit, as shown in the figure. Determine the amount of power dissipated in the three resistors.

Homework Equations



loop law
junction law

The Attempt at a Solution



(1) i1 +i2 +i3 = 0

(2) 9 - 45 i2 + 32 i1 = 0

(3) 15 +60 i3 + 9 - 45 i2 = 0

when you use the loop law, don't you have to go the same direction with every loop?
In the equation (3) it is going clockwise, and in equation (2) it is going counter-clockwise.
Is this wrong?
 
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You don't have to go the same direction with every loop. All you have to do is respect the potential drops/rises as specified by the chosen current directions (as indicated by the arrows) as you traverse the components during your "tour" around the loop.
 
Hi Cisneros778! :smile:
Cisneros778 said:
when you use the loop law, don't you have to go the same direction with every loop?
In the equation (3) it is going clockwise, and in equation (2) it is going counter-clockwise.
Is this wrong?

No, it doesn't matter which way round you go …

if you went the other way round one loop, everything would be multiplied by -1, wouldn't it? :wink:

btw, it might be easier to take the outside loop rather than the bottom loop (the outside loop is the sum of the other two), since it has one less term in it (and one less chance of making a mistake! :biggrin:)​
 
Thank you so much. I got the answers correct!
 

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