Above you see an LC tank combined of two ideal capacitor and one ideal inductance.

And general equations of condansators and inductance are as follows (when SW is closed):

$ \frac{d(V_1(t))}{dt}\times C = I(t) $

$ \frac{d(V_2(t))}{dt}\times C = -I(t) $

$ \frac{d(I(t))}{dt}\times L = V_1(t) - V_2(t) $

Now, think the initial condition as this:

SW open and left side capacitance charged up to $V_1$ which is higher than $V_2$. After capacitance on the left is charged SW is closed and see what happens. Since all the components are ideal and there is no loss I assumed that there will be an oscillation.

$ \frac{d^2(I(t))}{(dt)^2}\times L \times C = \frac{d(V_1(t))}{dt} \times C - \frac{d(V_2(t))}{dt} \times C $

$ \frac{d^2(I(t))}{(dt)^2}\times L \times C = 2 \times I(t)$

when I solve this differential equation:

$ I(t) = C_1 \times e^{-(\frac{\sqrt{2}t}{\sqrt{LC}})} + C_2 \times e^{(\frac{\sqrt{2}t}{\sqrt{LC}})} $

Now, I can think that $V_1, V_2, I$ relation will be something like this when SW is closed:


It is reasonable because when derivatives of $V_1(t)$ and $V_2(t)$ is zero (at local maxima and minima) $I(t)$ is zero and when they are highest $I(t)$ also highest. So I think it is reasonable to assume that at at t = 0, $I(t) = 0$ since I assumed that in initial state $V_1(t)$ was pulled the highest voltage and $V_2(t)$ was pulled down to lowest voltage level (local minima and maximas)

Lastly, I can also assume that since $I(t)$ starts as a sine wave, derivative of $I(t)$ is 1 at t = 0;

$ \frac{dsin(t)}{dt} = cos(t) $ , $cos(0) = 1$

$ \frac{dI(t)}{dt} = 1 $

When to finalize it I solve the equation with initial conditions but this is what I get:

$ I(t) = \frac{ \sqrt{LC} \times ( e^{ \frac{ 2 \sqrt{2} t}{ \sqrt{CL} } } - 1) }{2 \sqrt{2} e^{ \frac{ 2 \sqrt{2} t}{ \sqrt{CL} } } } $

But this function does not oscillate. It is something like this:


So could you please tell me where am I doing wrong and what is the correct equation?


You have an error with your signs in these equations.

$ \frac{d(V_1(t))}{dt}\times C = I(t) $

$ \frac{d(V_2(t))}{dt}\times C = -I(t) $

It is also not clear which sign convention you are using.

Given the direction of the current $I(t)$ as shown in your diagram

$ \frac{d(0-V_1(t))}{dt}C = I(t) $

$ \frac{d(V_2(t)-0)}{dt}C = I(t) $

where I have assumed that the current flows due to a potential difference across the component in a direction from higher potential to lower potential.

In terms of the equation $\frac{d(0-V_1(t))}{dt}C = I(t) $, initially the voltage across the left hand capacitor will be dropping so $\frac{dV_1(t)}{dt}$ will be negative and $-\frac{dV_1(t)}{dt}=I$ will be positive as you would expect?

Your differential equation then becomes

$ \frac{d^2I}{dt^2}= -\frac{2}{LC}I$

That negative sign makes all the difference as it gives a solution

$ I(t) = A_1 e^{-j(\frac{\sqrt{2}t}{\sqrt{LC}})} + A_2 e^{j(\frac{\sqrt{2}t}{\sqrt{LC}})} $ or $ I(t) = B_1 \sin (\frac{\sqrt{2}t}{\sqrt{LC}}) + B_2 \cos(\frac{\sqrt{2}t}{\sqrt{LC}}) $

which is the oscillatory solution you were looking for.


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