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# RC natural response - intuition

We put some charge on a capacitor in an RC circuit and observe what happens to the voltage and current. This is called the natural response. Created by Willy McAllister.

## Want to join the conversation?

• how the voltage across capacitor is vo? Some amount has to drop across resistor isn't it?
• With the switch in the initial position (connected to the battery) for a long time, a long time ago a current flowed that charged up the capacitor to Vo. Now consider the time in the recent past, just before we flip the switch. The voltage on both ends of the resistor is the same, Vo. The voltage across the resistor is Vo - Vo = 0 volts. Ohm's Law tells us the current through the resistor is i = V/r = 0/R = 0 amps. So just before we flip the switch, the initial conditions are Vc = Vo, Vr = 0, and i = 0.
• does frequency response and natural response the same thing?
• Hello Nural(a),

Not exactly.

Let's use the analogy of a wine glass. Ref: https://www.youtube.com/watch?v=BE827gwnnk4

The "natural response" term is easy. This is the frequency that the wine glass naturally "rings" at. In the video this is the frequency that ultimately destroys the glass. Know that the wine glass analogy is applicable to electronics circuits containing resistors, inductors, and capacitors. The difference is that the electronic circuits don't blow up... Usually...

"Frequency response" generally implies a chart that describes how the circuit responds to a broad range of frequencies. In the video this is like changing the frequency and noting how the glass responds to each frequency. For example, the glass does not respond to low frequencies but responds greatly to the natural frequency. One way to present this information is to use Bode plots. Ref

https://en.wikipedia.org/wiki/Bode_plot

Regards,

APD
• during discharge , the voltage across the capacitor will not be reversed?

and why does the charge on the upper plate will go to neutralize with the negative charge in the lower plate?
• With the RC circuit the voltage on the capacitor does not reverse (does not change sign). At the outset, the battery places a positive charge on the top capacitor plate. That means there is a charge difference between the top and bottom plate. That charge difference, q, causes a voltage to appear on the capacitor, v = q/C.

When you throw the switch the battery is removed from the circuit and there's a new path created that allows charge to flow from the top plate, through the resistor R, and onto the bottom plate.

Over time, there's less and less excess charge on the top plate, and more and more charge on the bottom plate. Eventually there is an identical amount of charge on both plates, which means there is no "excess" charge anywhere. When that happens the voltage on the capacitor falls to v = 0/C = 0.

At that point there's no more charge on the top plate that could possibly flow to the bottom plate and make it take on a negative voltage. The capacitor voltage never reverses.
• So do you have to review this to yourself or do you remember it all? If you do, that is impressive because I got like 30 something pages of notes..I can't remember everything but I am getting the basic concepts.