Calculate Battery Energy Delivery Rate for Parallel Circuit

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Homework Help Overview

The discussion revolves around calculating the energy delivery rate of a battery in a parallel circuit containing three resistors with specified resistances. The original poster presents a scenario where a current of 2 A flows through a 9-ohm resistor, and seeks to determine the energy delivery rate to the entire circuit.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expression for equivalent resistance in parallel circuits and the energy dissipation in resistors. There are inquiries about the applied voltage across the parallel combination and the calculations involved in determining the equivalent resistance.

Discussion Status

Some participants have provided calculations for the equivalent resistance and expressed understanding of the voltage across parallel branches. There is acknowledgment of a misstep in using the correct equation for energy dissipation, indicating a productive exchange of ideas.

Contextual Notes

Participants are working within the constraints of the problem statement, which specifies the current and resistance values but does not provide additional context or parameters for the circuit analysis.

Ruleski
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Homework Statement



A current I9 = 2 A flows through the 9 ohm resistor. At what rate does the battery deliver energy to the entire circuit?

Homework Equations


There are 3 resistors in the circuit, all wired in parallel. r1=15 ohms, r2=12 ohms, r3=9 ohms


The Attempt at a Solution

 
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What is the expression for the equivalent resistance of the parallel combination?

What is the energy dissipation in the resistance?

In the problem I9 and r3 is given. Find the applied voltage across the parallel combination.
 
All I know is that resistance is 1/total=1/r1+1/r2+1/r3. I calculated 3.83 ohms
 
Ruleski said:
All I know is that resistance is 1/total=1/r1+1/r2+1/r3. I calculated 3.83 ohms

In parallel combination voltage across each branch of the resistance is the same.

So V = I9*r3.

Energy dissipation is W = V^2/r.
 
Thank You. I was using the wrong equation for the last step
 

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