A 3.0 / Omega Ω resistor is connected in parallel to a 6.0 / Omega Ω resistor, and the combination is connected in series with a 4.0 / Omega Ω resistor. If this combination is connected across a 12.0 V battery, what is the power dissipated in the 3.0 / Omega Ω resistor?

The power dissipated in the 3 Ω resistor is P = 5.3watts.

Explanation:

After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.

The resultating resistor is of Req=6Ω.

I = V/Req

I = 2A

the parallel resistors have a potential drop of Vparallel=4 volts.

I (3Ω) = Vparallel/R (3Ω)

I (3Ω) = 1.33A

P = I (3Ω) ² * R (3Ω)

P = 5.3 Watts