# Power Dissipated by Resistor

1. Mar 9, 2009

### cwesto

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

How much power is dissipated by the 12$$\Omega$$ resistor in the figure?

PR1=______W
How much power is dissipated by the 18 resistor in the figure?
PR2=______W

2. Relevant equations

P=IV
I=$$\frac{V}{R}$$

3. The attempt at a solution

P=$$\frac{V}{R}$$*V
P=V2/R

P=12*12/12=12V

After solving for my equation I plugged in my numbers but it's not correct. All I need are a few hints to lead me in the right direction. Thanks!

Last edited: Mar 9, 2009
2. Mar 9, 2009

### hage567

Please show the details of your calculations so we can see what you're doing. You need to find the voltage drop across the resistor you're interested in. Note this will NOT be 12V.

3. Mar 9, 2009

### Kruum

Try finding either the current through the resistors or the voltage across the resistors.

4. Mar 9, 2009

### Ofey

What is the current in the circuit?

The power is

P=RI^2

Where R is the resistanse of the component and I is the current through the component

5. Mar 9, 2009

### cwesto

the current is 2 amps. what is the equation to find the voltage drop?

6. Mar 9, 2009

### Kruum

It's Ohm's law. But all you need is the current. Try arrangin the power equation so that you only need to use the current and resistor. Is that the current for the whole circuit? If so it's incorrect.

7. Mar 9, 2009

### cwesto

wait sorry they dont give the current my bad. i was looking at another problem. sorry
all the information up there is what they give me to solve the problem, nothing more.

8. Mar 9, 2009

### Kruum

You can find the current using the information in the problem. Okay, you know Ohm's law and you probably can create an Req for the two resistor in the circuit. That's should be enough to find the current.

9. Mar 29, 2010

### sweetdion

So I found the power in each resistor using P=I^2R, plugging in the current going through that resistor and the resistance and I get the sum of the power being 204 watts. This is not what I got in part c. Is it supposed to be?

10. Mar 29, 2010

### collinsmark

Hello sweetdion,

I think you might have posted to the wrong thread.

11. Mar 29, 2010

### sweetdion

Haha, yes, thank you.