Answer
a) $R_{eq}=6.5\Omega$
b) $I_{14}=0.429A$
c) $I_{12}=0.317A$
d) $P_{4.5}=1.06W$
Work Step by Step
a) The $12\Omega$ and $22\Omega$ are in parallel, so
$R_{eq1}=(\frac{1}{12\Omega}+\frac{1}{22\Omega})^{-1}=7.76\Omega$
These are in series with $4.5\Omega$
$R_{eq2}=7.76\Omega+4.5\Omega=12.3\Omega$
This is in parallel with $14\Omega$
$R_{eq}=(\frac{1}{12.3\Omega}+\frac{1}{14\Omega})^{-1}=6.5\Omega$
b) $6.0V-I_{14}(14\Omega)=0$
$I_{14}=0.429A$
c) $6.0V-I_{12}(12\Omega)-I_{4.5}(4.5\Omega)=0$
$6.0V-I_{22}(22\Omega)-I_{4.5}(4.5\Omega)=0$
$I_{12}(12\Omega)=I_{22}(22\Omega)$
$I_{12}=\frac{I_{22}(22\Omega)}{12\Omega}$
$I=\frac{V}{R_{eq}}=\frac{6.0V}{6.5\Omega}=0.92A$
$I=I_{14}+I_{12}+I_{22}$
$0.92A=0.429A+\frac{I_{22}(22\Omega)}{12\Omega}+I_{22}$
$I_{22}=0.17A$
$I_{12}=0.32A$
d) $6.0V-(0.32A)(12\Omega)-I_{4.5}(4.5\Omega)=0$
$I_{4.5}=0.48A$
$P_{4.5}=(I_{4.5})^2(4.5\Omega)=1.06W$