Answer
a) $V_{ad}=34.4V$
b) $V_{1}=82.3V$
$V_{2}=43.3V$
Work Step by Step
a) $I_3=I_1+I_2$
$45V=48\Omega (I_1+I_2) +34\Omega I_1$
$45V=82\Omega I_1+48\Omega I_2$
$I_1=\frac{45V-48\Omega I_2}{82\Omega}=\frac{45V-48\Omega I_2}{82\Omega}=0.549A-0.585I_2$
$45V+85V=48\Omega (I_3)+19\Omega I_2$
$130V=48\Omega (0.549A-0.585I_2)+67\Omega I_2$
$I_2=\frac{103.7V}{38.9\Omega}=2.67A$
$I_1=0.549A-0.585I_2=-1.01A$
$I_3=I_2+I_1=2.67A-1.01A=1.66A$
$V_{ad}=I_3R-\mathcal{E}_2=(1.66A)(48\Omega)-45V=34.4V$
b) $V_{ter}=\mathcal{E}-Ir$
$V_{1}=\mathcal{E}_1-I_2r=85V-(2.67A)(1\Omega)=82.3V$
$V_{2}=\mathcal{E}_1-I_3r=45V-(1.66A)(1\Omega)=43.3V$
