Charging a capacitor

A simple RC circuit with a DC power supply. When you close the switch, what happens?
K34.36

[@joeya5438]

I came to your most recent video just to tell you you are the man. Intelligent, attractive, charismatic, and just straight up contributing to the collective enlightenment of other humans through science. This man deserves maximum vindication for his efforts!

[@yoprofmatt]

Joey, Wow, now this is a flattering comment. Thank you very much. I'll do my best to keep the videos coming. Cheers, Dr. A

[@MohaMMaDiN55]

Thanks, Prof. Matt Anderson. You are really a professor You must be my teacher one day.

[@channel12381]

Semma🔥Explanation! Feeling bad because I don't have(can't find) a Physics Teacher in my school and college days

[@pratikshirsat1749]

Thank you sir this one helped for exams

[@yoprofmatt]

This is music to my ears. Cheers, Dr. A

[@tnekkc]

In discharge: V2/Vc = e^(-t/rc) In charging: V2/Vc = 1- e^(-t/rc) If I had a big schematic to design, these were fast, conditioned response. I did it for money at a lot of companies, designing power supplies and battery chargers from 1977 to 2008.

[@arahul4045]

Who else thinks he looks a bit like Tom Cruise???

[@koondoog4282]

For some reason this guy reminds me of Christopher Reeve/Benedict Cumberbach, anyone else?

[@yoprofmatt]

Haven't heard the Christopher Reeve likeness before. But I'll take it, thanks. Cheers, Dr. A

[@MVTchen01]

@@yoprofmatt o

[@kabandajamir9844]

So nice

[@JustNow42]

When the capacitor is charged through the resistor there will be an energy loss in the resistor. That loss is exactly half the energy in the capacitor when fully charged. It seems surprising that the loss does not depend on the resistors resistance. So what happens when the resistance is zero? An infinite current, a singularity? In real life we would get an oscillation because there always is a self-induction and energy would be radiated. It means that in a world where we assume there is no inductance while we have capacitors our theory must be incomplete.

[@khanahmad6354]

Very helpful sir

[@eng.khaledmoustafa8466]

WOW !

[@yoprofmatt]

Eng.Khaled Moustafa, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

[@franciscosanchez9406]

Does the capacitor pushes the amps that the load needs or more then it requires?

[@fransmurati2370]

OMG it is because of teacher like this that I ran.

[@fredlikestosmile2393]

am i the only one freaked out by him writing backwards, like so we can read it, not him??

[@santiagomaass5780]

they flipped the video, so no.

[@boboften9952]

WWW.LEARNING.GLASS Dr . MATT ANDERSON .

[@ronitmandal7301]

Seems like I am the first viewer 🎉 Thanks sir for these awesome videos

[@yoprofmatt]

Good to be number one! Thanks for tuning in. Cheers, Dr. A

[@sahilsagwekar]

Professor, if there is no flux accross the capacitors, so how does it 'conduct' electricity through it ?

[@yoprofmatt]

Sahil Sagwekar VEVO, Only works with AC current. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

[@franciscosanchez9406]

By what i mean a load max is 3 amps does it only pushes 3 amps or does the capacitor pushes more than 3 amps

[@yoprofmatt]

It is definitely limited by the resistive load. In our circuit, the current is limited by I0 = V / R as shown at 9:07 Cheers, Dr. A

[@mbregante2]

What happens when you charge a capacitor that has charge on it lesser than the potential of the battery?

[@isacdaimary9908]

I think they will level up until steady state is reached

[@yoprofmatt]

Then it will have a smaller voltage than the battery and not as much charge on it as it could have, since Q=CV. Cheers, Dr. A

[@juliarampelsberger7889]

👍👍

[@yoprofmatt]

Cool cool. Cheers, Dr. A