Deriving the Capacitor Discharge Equations

[@Robo2415]

Simple and straight to the point. That statement about relating the current to charge specifically on the capacitor was very well said. We need more videos like this!

[@te-kowski]

Great refresher for me. Love the brevity

[@iffatzarif7025]

Thank you so much! I finally understand where this formula comes from!

[@TheBrainFiller]

Great to hear thanks for watching!

[@swright4171]

Great video, thanks!

[@TheBrainFiller]

Thanks for watching!

[@yosefpatinio3669]

Good Video! suppose you only have the current equation . if you want to get Q (charge equation) , you do the integral , do you get the same Q as in the video? can you show it? i mean if you solve i(t)=i_0*exp[-t/(RC)] , i=dq/dt -> dq/dt = i_0*exp[-t/(RC)] -> q = Q_0*exp[-t/(RC)] + Q_0 (initial charge)

[@TheBrainFiller]

Yeah you get the same Q as in the video but just have to do a change of variables. When you integrate the exponential it brings down a factor of RC, so you get Q=(-I_0*RC)*exp(-t/RC)+c. Now at t=0 we want the charge to be Q_0 and at t=infinity we want the charge to be 0 (on the physical grounds that we believe this capacitor doesn’t have some fixed residual charge on it…we could of course have it be anything if we were just doing maths). So, the constant c has to be 0 so that it doesn’t decay to the constant c. We get the Q_0 by just looking at I_0*RC and relabelling it Q_0 (note as a sense check the units make sense: RC has units of time and I_0 has units of charge/time). Thanks for watching and hope that helped! Let me know if you want any clarification

[@yosefpatinio3669]

​@@TheBrainFiller Thanks for the answer , i appreciate it , " t=infinity we want the charge to be 0 " that helped me to understand ! btw i have another question , the charge equation (Q(t)) it have to be negative? i mean because of it is the electron...?

[@justlovegrov3549]

Thanks dawg appreciate it

[@IKdeoSSa_7]

This is good. I appreciate it.

[@TheBrainFiller]

Very kind, thanks for watching

[@AB-ts7gi]

Is there still a voltage drop across capacitor even though it acts as a Source of voltage?

[@te-kowski]

He derived the voltage. It is a function of time. If you go out to a large time, then no current is flowing, hence no voltage drop. During the time it acts as a source of voltage, then it isn't a voltage drop, rather a step up in voltage to complete kirchoffs voltage law of adding to zero in a loop. The positive voltage drop during the time the capacitor acts as a source of voltage is across the resistor.

[@R3ktByAGirll]

@@te-kowskithank you

[@Sparks09]

Nice one

[@TheBrainFiller]

Very kind, thanks for watching

[@grimeball]

I understand the logic of -dq/dt being negative because over a time interval there’s less charge flowing. But in the derivation of charging a capacitor, you use positive dq/dt. But in charging, over a time interval, less charge flows. Why is it positive in charging but negative here?

[@yosefpatinio3669]

in a charging capacitor , dq/dt is always positive , also when less charge flows too because the charge is only increasing! is it right that over the time less charge flow to it , but the charge is always growing until the capacitor reach is maximum voltage ( the same voltage of the battery) for example : at time t=1 dq/dt = 3 , at time t=2 dq/dt =1 and so... you see that dq/dt is less but still positive now when the capacitor is discharging , dq/dt is always negative for example : at time t=1 dq/dt = -5 , at time t=2 dq/dt = -2 and so... btw i didn't find a video of this channel about charging a capacitor....