Power dissipated in circuit combination

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Discussion Overview

The discussion revolves around the power dissipated in a circuit with three resistors, where R1 equals R2 and R2 equals two times R3. Participants explore the relationships between the powers dissipated in each resistor (P1, P2, and P3) and the total current in the circuit, utilizing various circuit laws and equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant attempts to find the total resistance and current in the circuit, expressing power dissipated in terms of the resistances and current.
  • Another participant suggests using Kirchhoff's law and current division to relate the currents through R2 and R3, proposing a simpler approach by assigning specific values to the resistors.
  • A participant derives expressions for the power dissipated in R1 and R2, concluding that P2 is one-ninth of P1, given R1 equals R2.
  • There is a question raised about the power dissipated in R3 (P3) and its relationship to P1.

Areas of Agreement / Disagreement

Participants generally agree on the relationships between the powers dissipated in R1 and R2, but the discussion about P3 remains unresolved, with no consensus on its value or relationship to P1 and P2.

Contextual Notes

Some assumptions about the circuit configuration and the values of the resistors are made, but these are not universally accepted or confirmed. The discussion includes various mathematical steps that are not fully resolved.

alex21
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Homework Statement



three resistors R1=R2 and R2=2R3, are connected to a battery, as show in the figure, if the power dissipated for each one is p1,p2 and p3 respectirvely;p3 must be equal to:

Homework Equations



P=VI
P=V^2/R
P=I^2*R

The Attempt at a Solution



I have tried to find the total resistance in the circuit by knowing that R2 and R3 are in parallalel and this equivalent resistance let's call it R23 will be in series with R1 so the equivalent resistance would be Req=R1+R23 this will allow me to find the total current in the circuit which is I=V/Req while doing so I should be able to find the power in R1 with this P=I^2*R1 since the total current in the circuit is the same going through R1, but I can't find the relationship with the power dissipated with P2 and as you can see P1 is in terms of P2
 

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Your third Relevant Equation is important. The question wants you to express the power dissipated by R1 in terms of the power dissipated by R2. So aim towards expressing R1 in terms of R2 and the current through R1 in terms of the current through R2. KCL and the current division rule will come in handy :wink:
 
alex21 said:

Homework Statement



three resistors R1=R2 and R2=2R3, are connected to a battery, as show in the figure, if the power dissipated for each one is p1,p2 and p3 respectirvely;p3 must be equal to:

Homework Equations



P=VI
P=V^2/R
P=I^2*R

The Attempt at a Solution



I have tried to find the total resistance in the circuit by knowing that R2 and R3 are in parallalel and this equivalent resistance let's call it R23 will be in series with R1 so the equivalent resistance would be Req=R1+R23 this will allow me to find the total current in the circuit which is I=V/Req while doing so I should be able to find the power in R1 with this P=I^2*R1 since the total current in the circuit is the same going through R1, but I can't find the relationship with the power dissipated with P2 and as you can see P1 is in terms of P2

Welcome to the PF.

Hint -- a simpler approach would be to assign values to the resistors (like 1 Ohm and 2 Ohms), and then look at how current divides between R2 and R3. Label the current through R2 = I. Then what is the current through R3? And what does that mean the current is through R1?

And solve...Please show us your work toward the solution! :smile:
 
lets call i the total current in the circuit therefore the current that goes through R1. applying kirchhoff's law to the junction where R2 and R3 splits i=i1+i2 the total current is the sum of the current that goes through R2 and R3
so we have:
i=i1+i2
i=V/R2 + V/R3
i=(R3+R2)V/R2*R3
but R2=2R3
i=(R3+2R3)V/R2*R3
i=3V/R2

I will assume V is the voltage drop on R2 so
V=iR2/3
And the power dissipated P2=V^2/R2 and P1=I^2*R1 so we have
P2=(iR2/3)^2/R2
P2=i^2R2^2/9R2
P2=i^2R2/9
but R1=R2 and P1=I^2*R1
P2=1P1/9
P1=9P2
 
Correct! :smile:
 
berkeman said:
Correct! :smile:

it would make any sense to ask P3?
 
alex21 said:
it would make any sense to ask P3?

For extra credit? Sure!

What is P1/P3?
 

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