Nodal analysis on AC RCL Circuit (I'm confused)
Assuming \$f=60\:\textrm{Hz}\$ and using the included schematic editor:







simulate this circuit &ndash; Schematic created using CircuitLab

From this, I get:

$$\begin{align*}
V_B &amp;= V_C + V_2\\\\
\frac{V_A}{4}+\frac{V_A}{3-2j}+\frac{V_A}{4j}&amp;=\frac{V_1}{4}+\frac{V_C}{3-2j}+\frac{V_B}{4j}\\\\
\frac{V_B}{4j}+\frac{V_B}{2j}&amp;=I_2+\frac{V_A}{4j}\\\\
\frac{V_C}{14}+\frac{V_C}{3-2j}+I_2&amp;=\frac{V_A}{3-2j}
\end{align*}$$

With \$V_1=100\angle 0^\circ\$ and \$V_2=220\angle 0^\circ\$, These solve out simultaneously as:

$$\begin{align*}
V_A&amp;=-28.9859544 - 29.8351871j\\
V_B&amp;=-1.19022489 + 96.0915807j\\
V_C&amp;=-221.190225 + 96.0915807j\\
I_2&amp;=79.5274823 - 6.35381992j
\end{align*}$$
or,
$$\begin{align*}
V_A&amp;\approx &amp;41&amp;.6\:\textrm{V}&amp; \angle -134.2&amp;^\circ\\
V_B&amp;\approx &amp;96&amp;.1\:\textrm{V}&amp; \angle +90.7&amp;^\circ\\
V_C&amp;\approx &amp;241&amp;.2\:\textrm{V}&amp; \angle +156.5&amp;^\circ\\
I_2&amp;\approx &amp;79&amp;.8\:\textrm{A}&amp; \angle -4.6&amp;^\circ
\end{align*}$$

Computing \$V_B-V_A\$ (the voltage across \$L_1\$) I get \$V_B-V_A\approx 129\:\textrm{V} \angle +77.553^\circ\$. If all these were RMS voltages, then this would in fact mean that the peak voltage across \$L_1\$ would be roughly \$182\:\textrm{V}\$.