Find $V_0$ in the circuit below.

Solution:

$KVL$:$(R_1+j2)\frac{*}{b}ar(I{)}_1+(j1)\frac{*}{b}ar(I{)}_2\frac{?}{b}=ar(V{)}_s$

$KVL$:$(j1)\frac{*}{b}ar(I{)}_1+(R_2+R_3+j2)\frac{*}{b}ar(I{)}_2=0\frac{?}{b}$

$ar(I{)}_1=\frac{3\mathrm{.}51}{-}37\mathrm{.}88A\frac{?}{b}$

$ar(I{)}_2=1\mathrm{.}24-172\mathrm{.}88A\frac{?}{b}$

$ar(V{)}_o\frac{?}{b}=ar(I{)}_2*R_3\frac{?}{b}$

$ar(V{)}_o=1\mathrm{.}24-172\mathrm{.}88V$

Explain how the currents and output voltage are calculated