# Homework Help: Simple Power Problem HELP!

1. Feb 6, 2008

### BuBbLeS01

1. The problem statement, all variables and given/known data

How much power is dissipated by R1? (R1=8 Ω, R2=17 Ω, V=6 V.)

Sorry the picture is so big!!!

2. Relevant equations

3. The attempt at a solution

P = E^2/R = 8^2/6 = 10.67W what am I doing wrong?

2. Feb 6, 2008

### BuBbLeS01

Do I have to add the resistors??

3. Feb 6, 2008

### Mthees08

No P=IV so the change in power is the dissipation of power so delta P=IV-IV
so use v=IR and get P=I^2R -I^2R for the points after and at (before) the resistor, also remember always final-initial

4. Feb 7, 2008

### BuBbLeS01

So I would do...
I = E/Req = 6/25
P = 0.24^2 * 8 - 0.24^2 * 17 = -0.5184
Negative? Why would it be negative?

5. Feb 7, 2008

### BuBbLeS01

Can someone explain this....

No P=IV so the change in power is the dissipation of power so delta P=IV-IV
so use v=IR and get P=I^2R -I^2R for the points after and at (before) the resistor, also remember always final-initial

I don't understand what Vi and Vf are?

6. Feb 7, 2008

### BuBbLeS01

so i need to multiply the voltage drop across the resistors by the current. but how do i calculate the voltage drop?

7. Feb 7, 2008

solved.....