We have two equations for the currents $I_1$ and $I_2$:
\[
\begin{aligned}
&L_{c} \frac{d I_{1}}{d t}+R I_{1}+2.5 I_{2}=E_{n} \\
&\frac{d I_{2}}{d t}+2.5 I_{1}-I_{2}=0
\end{aligned}
\]
where $L_{c}=1 \mathrm{H}, R=2.5 \Omega, E_{n}=845 \sin t \mathrm{V},
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