Calculating Power Dissipation in a Parallel Circuit with Multiple Resistors

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To calculate the power dissipation in the 8.0-ohm resistor within the given circuit, first determine the equivalent resistance of the resistors in series and parallel configurations. The 3-ohm resistor is in series with a combination of the 2-ohm and 1-ohm resistors, which should be added together first. After finding the total resistance, calculate the total current using Ohm's Law, then find the voltage across the 8-ohm resistor. Finally, the power dissipated in the 8-ohm resistor can be calculated by multiplying the current through it by the voltage across it.
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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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