A DC voltage source excites an RLC circuit, as shown. At time $t = 0$ , the inductors and capacitors are de-energized. The quantity $I_S$ is the current flowing out of the DC source. Let $I_{S0}$ be the value of $I_S$ at time $t = 0$ . Let $I_{S \infty}$ be the limiting value of $I_S$ as the elapsed time approaches infinity. Let $I_{S min}$ and $I_{S max}$ be the smallest and largest values of $I_S$ over all time.

Determine the value of $Q$ :

$\large{ Q = I_{S0} \,I_{Smin} \, (I_{Smax} - I_{S \infty})}$

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Details and Assumptions:
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**
1)
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All
$RLC$
values are
$1$
.
$V_S = 10$

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2)
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All four current values are positive

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3)
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The circuit shown is a modified version of a circuit from
this web page

The answer is 20.415.

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I will post my solution as well, since there are minor stylistic differences relative to the solution given by @Karan Chatrath . The state space representation is:

$V_{L1} = V_S - R_1 I_{L1} - V_{C3} = L_1 \dot{I}_{L1} \\ V_{L2} = V_{C3} + V_{C1} - R_3 I_{L2} = L_2 \dot{I}_{L2} \\ V_{L3} = V_{C3} = L_3 \dot{I}_{L3} \\ I_{R2} = \frac{V_S - (V_{C3} + V_{C1})}{R_2} \\ I_{C1} = I_{R2} - I_{L2} = C_1 \dot{V}_{C1} \\ I_{R4} = \frac{V_{C3} - V_{C2}}{R_4} \\ I_{C2} = I_{R4} = C_2 \dot{V}_{C2} \\ I_{C3} = I_{L1} + I_{C1} - I_{R4} - I_{L3} = C_3 \dot{V}_{C3}$

Isolate the time derivatives of state variables $(I_{L1}, I_{L2}, I_{L3}, V_{C1}, V_{C2}, V_{C3})$ and numerically integrate. Then the source current is:

$I_S = I_{L1} + I_{R2}$