RLC 6-17-2020

Charge on $C_1$ : $Q_1$

Charge on $C_2$ : $Q_2$

Source current through $V_1$ : $I_{V_1}$

Source current through $V_2$ : $I_{V_2}$

Initial conditions:

$Q_1(0) = Q_2(0) = I_L(0)=0$

Circuit Equations:

$I_{V_1} = I_L + I_{C_1}$ $I_{C_2} = I_L + I_{V_2}$ $\dot{Q}_1 = I_{C_1}$ $\dot{Q}_1 = I_{C_2}$ $-V_1 + RI_{V_1} + \frac{Q_1}{C_1}=0$ $-V_2 + RI_{V_2} + \frac{Q_2}{C_2}=0$ $L\dot{I}_L +\frac{Q_2}{C_2} = \frac{Q_1}{C_1}$

Laplace Transform on both sides:

$I_{V_1} (s)= I_L(s) + I_{C_1}(s)$ $I_{C_2}(s) = I_L(s) + I_{V_2}(s)$ $sQ_1(s) = I_{C_1}(s)$ $sQ_2(s) = I_{C_2}(s)$ $-\frac{V_1}{s}+ RI_{V_1}(s) + \frac{Q_1(s)}{C_1}=0$ $-\frac{V_1}{s}+ RI_{V_2}(s) + \frac{Q_2(s)}{C_2}=0$ $LsI_L +\frac{Q_2(s)}{C_2} = \frac{Q_1(s)}{C_1}$

Plugging in parameters and solving for $Q_1(s)$ :

$Q_1(s) = \frac{20\,\left(s^2+10\,s+15\right)}{s\,\left(s+10\right)\,\left(s^2+10\,s+20\right)}$

Therefore voltage across capacitor $C_1$ :

$V_{C_1}(s) = \frac{Q_1(s)}{C_1}= \frac{200\,\left(s^2+10\,s+15\right)}{s\,\left(s+10\right)\,\left(s^2+10\,s+20\right)}$

Using an online inverse Laplace transform calculator and re-arranging the result a little, gives:

$\boxed{V_{C_1}(t) = 10\,\sqrt{5}\,{\mathrm{e}}^{-5\,t}\,\mathrm{sinh}\left(\sqrt{5}\,t\right)-15\,{\mathrm{e}}^{-10\,t}+15}$

Variation of $V_{C_1}$ with time:

I have solved the problem by just writing expression and solving differential equation , after this using graph.
But I am trying something new, I will post my solution using python code.