Hmmm power in a parallel/series circuit

[kpangrace]

Determine the power dissipated in the 9.0 resistor in the circuit shown in the drawing. (R1 = 3.0 , R2 = 9.0 and V1 = 12 V.)

ok here's my problem... I've found the current throughout to be 2.261 a...

now if i use that to find the voltage in the 9 ohm resistor, i get a voltage bigger than my batteries voltage!

i'm completely confused on where to go with that current to get to power...

any help on what exactly my next step is would be appreciated. thank you in advance!:shy:

 

[Diane_]

I'm getting a much smaller current. How did you get that value?

 

[kpangrace]

9+1= 10 ohm
2+1= 3 ohm

1/10 + 1/3= 1/r=2.308
2.308+3= 5.308 ohms
v=12
I= V/R

I= 12/5.308
I=2.261

did i not add the resistors correctly?

 

[Diane_]

I'm sorry - I missed the word "throughout" in your original post. That is the value I have for the total current in the circuit. It's the current through R2 for which I have a smaller value. Ah! And the light dawns over her head. You do realize that the total current will not be what goes through the 9 ohm resistor, right? You have two branches there - the top one with an effective resistance much smaller than the bottom one, so relatively little of the current will go through the bottom one. Do you see how to divide the current up between the two branches?

 

[kpangrace]

would i just do it as I=12/9 = 1.333 to get the current running through it? i assumed that because htere's both parallel and series circuits that they first wanted me to combine them all to practice that or whatever... hmmm the first problem on my homework page and the last one i need

 

[Diane_]

General principles:

1) In a purely series circuit, the current is the same through all elements. The voltage drop will differ, one from another.

2) In a purely parallel circuit, the voltage drop is the same across all elements. The current will differ.

You can find the voltage drop across R1. That will tell you the voltage drop that remains across the parallel portion of the circuit, which means the voltage drop across both the upper and the lower branch. Knowing that, you can find the current through the lower branch, which tells you the current through the nine ohm resistor.

There are quicker ways to get it, but they all stem from those basic ideas.

Is that sufficient?