Inductor and Capacitance (07-09-2020)

See the diagram above for the polarity and naming conventions. The state variables are the capacitor voltages $V_{C1}$ and $V_{C2}$ , and the inductor current $I_L$ .

The first equation is for the inductor.

$V_L = L \dot{I}_L \\ E + V_{C1} = L \dot{I}_L$

The next equation relates two different expressions for the voltage across the inductor:

$E + V_{C1} = 2 E - V_{C2}$

Taking the time derivatives of both sides yields an expression in terms of the time derivatives of the capacitor voltages.

$\dot{V}_{C1} = - \dot{V}_{C2} \\ \dot{V}_{C1} + \dot{V}_{C2} = 0$

Finally, the sum of the capacitor currents (expressed in terms of the capacitor voltages) equals the inductor current. Note that due to the polarity convention used, the Capacitor $1$ current is negated.

$I_{C2} - I_{C1} = I_L \\ C_2 \dot{V}_{C2} - C_1 \dot{V}_{C1} = I_L$

To summarize:

$E + V_{C1} = L \dot{I}_L \\ \dot{V}_{C1} + \dot{V}_{C2} = 0 \\ C_2 \dot{V}_{C2} - C_1 \dot{V}_{C1} = I_L$

In matrix form (note that $C_1$ and $C_2$ are equal to $C$ ):

\[\begin{bmatrix} L & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -C_1 & C_2 \end{bmatrix}

\begin{bmatrix} \dot{I}_L \\ \dot{V}_{C1} \\ \dot{V}_{C2} \end{bmatrix}

= \begin{bmatrix} E + V_{C1} \\ 0 \\ I_L \end{bmatrix} \]

This is a state-space representation in which the time derivatives of the state variables are expressed in terms of the state variables and the forcing functions (the voltage sources, in this case). One could try and solve this system of differential equations analytically. I used numerical integration instead.

Since the overall voltage across both capacitors in series is initially $E$ , initialize the state variables as follows:

$V_{C1} (0) = \frac{E}{2} \\ V_{C2} (0) = \frac{E}{2} \\ I_L (0) = 0$

The constant $\alpha \approx 2.121$ .

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import math
import numpy as np
import random

#################################################

# Constants

dt = 10.0**(-5.0)

C = 1.0
L = 1.0
E = 1.0

C1 = C
C2 = C

#################################################

# System Matrix

J11 = L
J12 = 0.0
J13 = 0.0

J21 = 0.0
J22 = 1.0
J23 = 1.0

J31 = 0.0
J32 = -C1
J33 = C2

J =  np.array([[J11,J12,J13],[J21,J22,J23],[J31,J32,J33]])

#################################################

# Numerical Simulation

t = 0.0
count = 0

IL = 0.0
VC1 = E/2.0
VC2 = E/2.0

ILd = 0.0
VC1d = 0.0
VC2d = 0.0

ILmax = 0.0

while t <= 20.0:

    IL = IL + ILd*dt
    VC1 = VC1 + VC1d*dt
    VC2 = VC2 + VC2d*dt

    vec = np.array([E+VC1,0.0,IL])

    Sol = np.linalg.solve(J, vec)

    ILd = Sol[0]
    VC1d = Sol[1]
    VC2d = Sol[2]

    if IL > ILmax:
        ILmax = IL

    t = t + dt
    count = count + 1

    #if count % 1000 == 0:
        #print t,IL

#################################################

# Print results

print ""
print ""

print dt
print t
print ILmax

#1e-05
#10.0000099998
#2.1213321245714

I was trying to solve using the concepts of energy and finally it worked
When the switch is open,Energy residing in the capacitors $E_{initial}=\frac{1}{2}C(\frac{E}{2})^{2}+\frac{1}{2}C(\frac{E}{2})^{2}$ $E_{initial}=\frac{1}{4}CE^{2}$ After the switch is closed,Energy residing in the capacitors $E_{final}=\frac{1}{2}C(E)^{2}+\frac{1}{2}C(2E)^{2}$ $E_{final}=\frac{5}{2}CE^{2}$
Energy supplied by the batteries $E_{batteries}=(\frac{3CE}{2}) E+(\frac{3CE}{2})2E$ $E_{batteries}=\frac{9}{2}CE^{2}$ $E_{initial}+E_{batteries}-E_{final}=\frac{1}{2}Li_{max}^{2}$
Solving the equation gives $\boxed{i_{max}=\frac{3}{\sqrt{2}}E\sqrt{\frac{C}{L}}}$

Current through right capacitor $I_2$ and that through left capacitor is $I_1$ . Charge on left capacitor is $Q_1$ and that through right capacitor is $Q_2$ . CUrrent through coil is $I$ .

Circuit equations:

$-E + \frac{Q_1}{C} + L \dot{I} =0$ $-2E + \frac{Q_2}{C} + L \dot{I} =0$ $I = I_1 + I_2$ $\dot{Q}_1 = I_1$ $\dot{Q}_2 = I_2$

Let $I(0)=Q_1(0)=Q_2(0)=0$

Laplace transform on both sides:

$-\frac{E}{s} + \frac{Q_1(s)}{C} + LsI(s)=0$ $-\frac{2E}{s} + \frac{Q_2(s)}{C} + LsI(s)=0$ $I(s) = I_1(s) + I_2(s)$ $sQ_1(s) = I_1(s)$ $sQ_2(s) = I_2(s)$

Solving for $I(s)$ gives:

$I(s) = \frac{3}{\sqrt{2}}E \sqrt{\frac{C}{L}}\left(\frac{\sqrt{\frac{1}{2LC}}}{s^2 + \frac{1}{2LC}}\right)$

Taking inverse Laplace transform gives:

$I(t) = \frac{3}{\sqrt{2}}E \sqrt{\frac{C}{L}} \sin\left(\frac{t}{\sqrt{2LC}}\right)$

Therefore:

$I_{max} = \frac{3}{\sqrt{2}}E \sqrt{\frac{C}{L}}$