A3.0 \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

are you like autistic or something?

xd

i think the answer is a, hope this is right, that would be my best guess.