Power dissipated by a Resistor

[andquotethat]

Homework Statement

What power is dissipated by the R2=3.0Ω resistor in the figure if R1=6.0Ω?
Battery 1: 12 V
Battery 2: 15 V
Resistor 1: 3.0 Ω
Resistor 2: 6.0 Ω

Homework Equations

P = V²/R

R_total = R₁ + R₂

The Attempt at a Solution

I started on the left side and used the formula, P = V²/R. I plugged 12 in for V and 3 for R. That resulted in 24 W from the battery on the left. I used the same formula on the right side but plugged in 15 for V and 3 for R. That resulted in 75 W. I added the 2 together and got 99 W.
That was the wrong answer.

I also tried adding 12 and 15 together for V and dividing it by the same R, 3. That gave me 240 W which was also an incorrect answer.

I tried to ratio it as well by taking the left side and 12²/9 then taking that result and dividing it by 3. I then did essentially the same to the other side but with 15 for the V and then taking a ratio of that as well. That resulted in 80 W. Once again, the answer was incorrect.

Am I even using the right equation?

 

[kuruman]

Are you familiar with Kirchhoff's circuit rules? Even if you are, check out this
https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
The example is very similar to your problem.

 

[Orodruin]

I started on the left side and used the formula, P = V2/R. I plugged 12 in for V and 3 for R.

To add to the previous reply: You cannot pick a formula and randomly start inserting numbers in it, that will typically never end well. Instead, try to understand what the components are.

In this case, V is the difference in potential across the resistor and R is the resistance of the resistor. You also cannot compute the contributions from each battery separately and add them since your equation is non-linear. You need to find the actual potential difference across the resistor using, eg, what was said in #2.

 

[andquotethat]

Thank you both. Most of the other problems I had to do were relatively simple. Understood however. I will step back next time and rethink the problem.

So I would make two loops then? The one on the left going clockwise inside the box and the other going clockwise inside the right box?

Then the current follows both loops but goes from the south to north position in the middle because the volts from the 15 V battery are greater than the volts from the 12 V battery?

Am I right so far?

 

[Orodruin]

Then the current follows both loops but goes from the south to north position in the middle because the volts from the 15 V battery are greater than the volts from the 12 V battery?

You would make two loops, yes. Note that each loop will carry an independent current. The current through the middle resistor will be given by the sum of the contributions from each loop (taken with the appropriate sign!).

 

[andquotethat]

Noted. I did finally figure out how to get the right answer. Thank you all for your help and I'll remember not to plug and chug next time.