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## Homework Statement

We are given a circuit as follows: V= 10 V and two resistors are in parallel. R1 is 12 ohms and R2 is 5 ohms. What is the power dissipated in the 12 ohm resistor?

## Homework Equations

V=IR

P=(I^2)R

## The Attempt at a Solution

I know that the total resistance of the circuit is 3.54 ohms... I found this by solving (1/Rtot) = (1/12)+(1/5) .

The total current of the circuit is 2.82 A, which I found by using V=IR, or 10 = (3.54) I. I will call this current I3.

I will just say that the current going through R1 is named I1 and the current going through R2 is named I2. Using V = IR, I can then say that 10 = (R1*I1) + (R2*I2). I can also say that I1 + I2 = 2.82 A.

I also know that P= V^2 / R. I can deduce that the total power dissipated is (10)^2/3.54, which is 28.2 W.

If the total power dissipated is the sum of each resistor's power dissipation, then 28.2 = (I1^2)(R1) + (I2^2)(R2).

I have a lot of pieces but I still am not sure how to figure out the power dissipated through a single resistor, especially since they have different resistances. Any guidance on where to go next?