Answer
(a) $192\Omega$
(b) $96\Omega$
(c) $150W$
Work Step by Step
(a) We can find $R_1$ as follows:
$R_1=\frac{V^2}{P_1}$
We plug in the known values to obtain:
$R_1=\frac{(120)^2}{75W}$
$R_1=192\Omega$
(b) We can find $R_2$ as follows:
As $R_1$ and $R_2$ are in series so $R_{eq}=R_1+R_2$
$\implies R_{eq}=\frac{V^2}{P_2}$
$\implies R_1+ R_2=\frac{V^2}{P_2}$
$\implies R_2=\frac{V^2}{P_2}-R_1$
We plug in the known values to obtain:
$R_2=\frac{(120V)^2}{50W}-192\Omega$
$R_2=96\Omega$
(c) We can find the required power consumed as
$P=\frac{V^2}{R_2}$
We plug in the known values to obtain:
$P=\frac{(120V)^2}{96\Omega}$
$P=150W$
