Alternating Current Series #1

The system equations are:

$E - V_C = L \, \dot{I}_{L1} \\ V_C = L \, \dot{I}_{L2} \\ I_{L1} - I_{L2} = C \dot{V}_C$

Subtracting the second equation from the first:

$E - 2 V_C = L \, (\dot{I}_{L1} - \dot{I}_{L2}) = L C \ddot{V}_C$

Homogeneous equation:

$-2 V_C = L C \ddot{V}_C$

This corresponds to simple harmonic motion with angular frequency $\omega = \sqrt{\frac{2}{L C}}$ . The general form is:

$V_C = A \cos(\omega t) + B \sin(\omega t) + D$

Apply the following initial conditions to solve for $(A,B,D)$ :

$V_C (0) = 0 \\ \dot{V}_C (0) = 0 \\ \ddot{V}_C (0) = \frac{E}{LC}$

After solving for the constants, the final answer is:

$V_C = \frac{E}{2} \Big(1 - cos(\omega t ) \Big) \\ \omega = \sqrt{\frac{2}{L C}}$

I have also included numerical simulation code:

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import math

E = 1.0
L1 = 1.0
L2 = 1.0
C = 1.0

dt = 10.0**(-6.0)

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

t = 0.0
count = 0

IL1 = 0.0
IL2 = 0.0
VC = 0.0

IL1d = 0.0
IL2d = 0.0
VCd = 0.0

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

while t <= 10.0:

    IL1 = IL1 + IL1d*dt
    IL2 = IL2 + IL2d*dt
    VC = VC + VCd*dt

    IL1d = (E-VC)/L1
    IL2d = VC/L2
    VCd = (IL1-IL2)/C

    t = t + dt
    count = count + 1

    if count % 10000 == 0:
        print t,VC

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

Let the currents through the inductor and the capacitor of the parallel circuit be $I_L$ and $I_C$ respectively. Then the voltage on the capacitor:

$\begin{aligned} V(t) & = V(0) + \int_0^t \frac {I_C(\tau)}C d\tau & \small \blue{\text{Since the capacitor is uncharged}} \\ & = \int_0^t \frac {I_C(\tau)}C d\tau & \small \blue{\text{at }t = 0 \implies V(0) = 0} \\ \frac {dV(t)}{dt} & = \frac {I_C(t)}C \\ \implies I_C(t) & = C \frac {dV(t)}{dt} \end{aligned}$

Note that the voltage on the capacitor is the same as the voltage of the parallel inductor. Then $V(t) = L\dfrac {dI_L(t)}{dt}$ . We also note that:

$\begin{aligned} L \frac d{dt} \left(I_L(t) + I_C(t)\right) + V(t) & = E \\ \blue{L\frac {dI_L(t)}{dt}} + LC \frac {d^2 V(t)}{dt^2} + V(t) & = E & \small \blue{\text{Note that }V(t) = L \frac {dI_L(t)}{dt}} \\ LC \frac {d^2 V(t)}{dt^2} + 2 V(t) & = E \end{aligned}$

Solving the differential equation (see Note):

$\begin{aligned} \implies V(t) & = c_1 \sin \left(\sqrt{\frac 2{CL}}t\right) + c_2 \cos \left(\sqrt{\frac 2{CL}}t\right) + \frac E2 & \small \blue{\text{where }c_1 \text{ and }c_2 \text{ are constants.}} \\ I_C(t) & = C \frac {dV(t)}{dt} \\ & = \sqrt{\frac {2C}L} \left(c_1\cos \left(\sqrt{\frac 2{CL}}t\right) - \sin \left(\sqrt{\frac 2{CL}}t\right)\right) \\ I_C(0) & = c_1 \sqrt{\frac {2C}L} = 0 & \small \blue{\implies c_1 = 0} \\ \implies V(t) & = c_2 \cos \left(\sqrt{\frac 2{CL}}t\right) + \frac E2 \\ V(0) & = c_2 + \frac E2 = 0 & \small \blue{\implies c_2 = - \frac E2} \\ \implies V(t) & = \frac E2 \left(1-\cos \left(\sqrt{\frac 2{CL}}t\right)\right) \end{aligned}$

Therefore $\alpha + \beta = 2+2 = \boxed 4$ .

Note : Consider a differential equation $\dfrac {d^2v(t)}{dt^2} + pv(t) = q$ . Then the characteristic equation $(D^2 + q)v = 0$ , $\implies D = \pm i\sqrt p$ and the general solution of the differential equation is:

$\begin{aligned} v(t) & = c_1 \sin (\sqrt p t) + c_2 \cos (\sqrt p t) + c_3 \\ \frac {dv(t)}{dt} & = \sqrt p \left(c_1 \cos (\sqrt p t) - c_2 \sin (\sqrt p t)\right) \\ \frac {d^2v(t)}{dt^2} & = - p \left(c_1 \sin (\sqrt p t) + c_2 \cos (\sqrt p t)\right) \\ & = - pv(t) + pc_3 \\ \implies \frac {d^2v(t)}{dt^2} + pv(t) & = q \\ - pv(t) + pc_3 + pv(t) & = q \\ \implies c_3 = \frac qp \end{aligned}$

Putting $v(t) = V(t)$ , $p = \dfrac 2{LC}$ , and $q = \dfrac E{LC}$ , then $V(t) = c_1 \sin \left(\sqrt{\frac 2{LC}}t\right) + c_1 \cos \left(\sqrt{\frac 2{LC}}t\right) + \dfrac E2$ .