The circuit below consists of three capacitors $(C_1 = 2, C_2 = 2, C_3 = 3)$ and three inductors $(L_1 = 2, L_2 = 3, L_3 = 1)$ . At time $t = 0$ the inductor currents are zero and the capacitor voltages are $(V_{C1} = +5, V_{C2} = +7, V_{C3} = -12)$ .

At time $t = 4$ , what fraction of the total system energy is contained within the three capacitors combined?

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Note:
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In the drawing, the plus and minus signs indicate voltage polarities, and the arrows indicate current polarities.

The answer is 0.15.

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Slightly different approach from that of @Hosam Hajjir

The circuit equation by applying Kirchoff's current and voltage laws read:

$\dot{Q}_3 = I_{C3}$ $\dot{Q}_2 = I_{C2}$ $\dot{Q}_1 = I_{C1}$

$I_{C3} = I_{C2} + I_{L3}$ $I_{C1} = I_{C3} + I_{L1}$ $I_{C2} = I_{C1} + I_{L2}$

$\frac{Q_3}{C_3} + L_3 \dot{I}_{L3} = L_1 \dot{I}_{L1}$ $\frac{Q_2}{C_2} + L_2 \dot{I}_{L2} = L_3 \dot{I}_{L3}$ $\frac{Q_1}{C_1} + L_1 \dot{I}_{L1} = L_2 \dot{I}_{L2}$

Taking Laplace transform on both sides of all equation leads to the following linear equations in the s-domain:

$sQ_3(s) - C_3V_{C3} = I_{C3}(s)$ $sQ_2(s) - C_2V_{C2} = I_{C2}(s)$ $sQ_1(s) - C_1V_{C1} = I_{C1}(s)$

$I_{C3}(s) = I_{C2}(s) + I_{L3}(s)$ $I_{C1}(s) = I_{C3}(s) + I_{L1}(s)$ $I_{C2}(s) = I_{C1}(s) + I_{L2}(s)$

$\frac{Q_3(s)}{C_3}+ sL_3 I_{L3}(s)=sL_1 I_{L1}(s)$ $\frac{Q_2(s)}{C_2}+ sL_2 I_{L2}(s)=sL_3 I_{L3}(s)$ $\frac{Q_1(s)}{C_1}+ sL_1 I_{L1}(s)=sL_2 I_{L2}(s)$

Plugging in all constants and solving for $Q_1(s)$ , $Q_2(s)$ and $Q_3(s)$ gives:

$Q_1(s) = \frac{160\,s}{16\,s^2+1}$ $Q_2(s) = \frac{4\,s\,\left(616\,s^2+31\right)}{\left(11\,s^2+1\right)\,\left(16\,s^2+1\right)}$ $Q_3(s) = -\frac{6\,s\,\left(1056\,s^2+71\right)}{\left(11\,s^2+1\right)\,\left(16\,s^2+1\right)}$

Using an online inverse Laplace transform calculator:

$Q_1(t) = 10\,\cos\left(\frac{t}{4}\right)$ $Q_2(t) = 20\,\cos\left(\frac{\sqrt{11}\,t}{11}\right)-6\,\cos\left(\frac{t}{4}\right)$ $Q_3(t) = -6\,\cos\left(\frac{t}{4}\right)-30\,\cos\left(\frac{\sqrt{11}\,t}{11}\right)$

The energy stored in the capacitors at any instant is:

$E_C(t) = \frac{Q_1^2}{2C_1} + \frac{Q_2^2}{2C_2} + \frac{Q_3^2}{2C_3}$

At time $t=0$ there is no current flowing through inductors, this means that the total energy of the system at that instant is stored in the capacitors. Therefore, the required answer is:

$\boxed{\frac{E_C(4)}{E_C(0)} = 0.15}$