Answer
a) $R_2=6.0\Omega$
b) $V_T=2.2V$
Work Step by Step
a) $R_{eq}=14.5\Omega+R_2$
$I=\frac{V}{R_{eq}}=\frac{12V}{14.5\Omega+R_2}$
$12.0V-\frac{(12V)(14.5\Omega)}{14.5\Omega+R_2}=3.5V$
$R_2=6.0\Omega$
b) $R_{eq}=14.5\Omega+\Big(\frac{1}{\frac{1}{6.0\Omega}+\frac{1}{7.0\Omega}}\Big)^{-1}=17.7\Omega$
$I=\frac{12.0V}{17.7\Omega}=0.677A$
$V_T=12.0V-(14.5\Omega)(0.677A)=2.2V$
