Answer
The power dissipated by the the $12~\Omega$ resistor is $1.92~W$
The power dissipated by the the $18~\Omega$ resistor is $2.88~W$
Work Step by Step
We can find the equivalent resistance of the two resistors in series:
$R = 12~\Omega + 18~\Omega = 30~\Omega$
We can find the current in the circuit:
$I = \frac{V}{R} = \frac{12~V}{30~\Omega} = 0.40~A$
We can find the power dissipated by the the $12~\Omega$ resistor:
$P = I^2~R = (0.40~A)^2(12~\Omega) = 1.92~W$
We can find the power dissipated by the the $18~\Omega$ resistor:
$P = I^2~R = (0.40~A)^2(18~\Omega) = 2.88~W$
