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# LC natural response derivation 1

We begin the derivation of the natural response of the LC circuit, by modeling it with a 2nd-order differential equation. Created by Willy McAllister.

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• Is it correct to say that in this LC circuit, the capacitor is also the voltage source? Would it be different if we have a separate voltage source instead?
• Hello Emmen,

Both the inductor and the capacitor will appear to be the voltage source. However, they never do so at the same time. Think of this as an energy storage problem. The energy constantly sloshes back and forth between the two devices.

In radio terms we call this a tank circuit as in water sloshing back and forth in a tank. The rate of energy exchange is known as the resonant frequency. The LC circuit is often found in radio frequency system as part of the station tuning mechanism.

Things get really interesting when you add separate voltage sources to this LC circuit. In general the LC circuit loves voltage sources that operate at the natural frequency. It will try to short out anything else.

Regards,

APD
• Hello ,
I have a problem understanding why the voltage on the capacitor and the inductor is the same , at least it shouldn't be the same just as you close the switch .
Any help ?
• Hello Hazem,

Before time zero the capacitor was charged to some voltage and the inductor is "empty" which is to say, no current is flowing in the inductor. At time zero the switch is closed. Instantaneously the voltage on both devices is equal. The voltage then oscillates as Willy describes in the video.

Recall that the voltage on an inductor is defined as:

V = Ldi/dt

This allows an instantaneous voltage jump when the switch is closed as the change in current is instantaneous (square wave).

Or stated another way, the inductor likes to keep current constant. Before the switch was closed the current was zero. Immediately after the switch is closed the inductor current is still zero. The only way this can happen is if the inductor's voltage instantaneously rises to the capacitor voltage.

Regards,

APD
• At Why is the current direction "reversed"? What is the reason for the minus sign in the equation?
• The negative sign is a small accident of how i is defined in the main LC circuit diagram on the left.

At I draw a capacitor over on the far right. The i-v equation for a capacitor is defined assuming the current, i_c, is coming down into the positive voltage terminal of the cap. With this current direction the capacitor equation is i_c = C dv/dt.

Now look at the direction of i_withnosubscript in the LC circuit on the left side of the screen. Notice it points in the opposite direction of i_c. It is flowing up out of the positive terminal of the cap. That means -i = i_c.

In the main drawing on the left there is no variable called i_c, there is just i. To write the capacitor i-v relationship correctly using i as the current variable you substitute in -i in place of i_c. This is how the negative sign gets introduced.
• Hello,
At , shouldn't the expression for voltage on the capacitor include a constant? (shouldn't we add the initial voltage on the capacitor?)
(1 vote)
• You can include an initial voltage on the capacitor if you want (and an initial current in the inductor. For this video I assumed both were 0, just for simplicity's sake.

Once you have gone through this solution it is interesting to go back and include those initial conditions on L and C. You will find that you get the same sinusoidal response but with a phase shift (the sine wave will start a little later or earlier than the 0 initial condition LC.