Calculating Power Dissipation in a Parallel Circuit with Multiple Resistors

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Homework Help Overview

The discussion revolves around calculating power dissipation in a parallel circuit involving multiple resistors, specifically focusing on an 8.0 ohm resistor in a circuit with a 10V battery and a 3.0 ohm resistor in series with several resistors in parallel.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for calculating equivalent resistance in series and parallel configurations, with some attempting to find total current and voltage across specific resistors. Questions arise regarding the correct order of operations and the handling of series and parallel combinations.

Discussion Status

Participants are actively engaging in exploring different approaches to the problem. Some have provided partial calculations and are questioning their methods, while others are clarifying the arrangement of resistors and the steps needed to find equivalent values. There is no explicit consensus on the correct approach yet.

Contextual Notes

Some participants express confusion over the arrangement of resistors and the calculations needed to find total current and power dissipation, indicating a need for further clarification on the setup and assumptions made in the problem.

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Determine the power dissipated in the 8.0 resistor in the circuit shown in the drawing. (R1 = 3.0 , R2 = 8.0 and V1 = 10 V.)

in the picture a 3 ohm resistor is connected in circuit and resistors 2ohms, 1ohm, 1ohm, and 8ohms are all connected in parallel to each other. all are connected to a 10v battery ( i think i attached the picture )
 

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ok here is what i tried...
i found the total I to be .381 i did that by adding 1/2+1/1+1/1+1/8=1/2.625=.381
next, i found the voltage after the 3ohm resistor. i did that by V=IR (.381)(3ohms)=1.143
then i found the current at the 8ohm resistor. I=v/r (10/3)=3.3
then, since the voltage across the parallel is the same i found the power at the 8ohm resistor P=VI (1.143)(3.3)=3.81
but that is wrong. where did i mess up?
 
Do all series equivalents.
Then do all parallel equivalents.
If, as an intermediate step, you wind up two in series, do that equivalent as soon as you can.
Keep units in your calculations.
 
so i add up 3 2 and 1 then 8 and 1 and get two resistors in parallel (6 and 9 ohms). then do i use the same equations as i did before?
 
No, I should been more specific and said simple series. The 3 Ohm is in series with the network not just the 2 Ohm and 1 Ohm. Try again.
 
so would i add up the resistors in parallel so that i ended up with two resistors in circuit first then add the two in circuit?
 
No. You have a 2 Ohm and a 1 Ohm in simple series. Combine them. What do you get? Are there any others in simple series?
 
8 and 1?
 
Yep. So now you have a 1 Ohm and a 2 Ohm combined in simple series. What is their equivalent value?

And, you have a 8 Ohm and a 1 Ohm combined in a simple series. What is their equivalent value.

How are the two equivalents arranged (series or parallel)? How would you combine them?
 
  • #10
3 ohm and 9 ohm connected in parallel. i add them 1/3+1/9=1/.444=2.25ohms. then i would add the other 3 ohm resistor to that giving me 5.25. now i have only one resistor. now i can find total I. right?
 
  • #11
You got it. Good work!
 
  • #12
Hey, I am having trouble with this almost identical question. What do you do after you find the total current?

You have to find the current that is traveling through the 8ohm resistor afterward right? How would I do that.

Then to get power it would just be the current traveling through the 8ohm resistor times the (total voltage - voltage of 3ohm reistor)?

Help please, very urgent!
 

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