Sorry for the dumb question sir, but this is just out of my curiosity, why the conservation of charge is applicable to only the inner part between 2 capacitor, not the outer part
Not a dumb question at all. The "why" questions are very important. In the case of a capacitor, the plates always have opposite charges. Therefore the charges on one plate attract the charges on the opposite plate, which causes the charges to migrate to the inner sides of the two plates.
4:20 The "initial" total charge is ZERO, not q1. It is only q1 if you consider the electrons alone (and they are the only moving charges in a solid). NOTHING forbids them to recombined with an atom momentarily missing an electron (a positive charge relevant to the discussion), or a neutral atom to generate an extra free electron , like they did to generate "q1" in the first place. So, I fail to see the support of the assertion about the "conservation" of the initial amount of negative charge. I agree that we must not destroy electrons, but the "free" ones can recombine and new free ones can be generated. Keeping the total energy, though, seems more fundamentally grounded, but that leads to a different result.
That is Kirchhoff's voltage law.. you can look it up.. It says that the algebraic sum of all voltages in a closed loop is 0.. It is applicable for all components regardless of their configuration.. hope this helps :)
Since the charges in both the capacitors become equal, the energy loss of the charged capacitor is same as energy gained by the uncharged one. Is that right?
OK, I tried this with different capacitors and initial charges on C1 with identical C2 discharged. I ALWAYS get a SMALLER NUMBER!! One reason I think is, is that when the two caps are connected together (very carefully, with one lead already connected, while the other is connected with a simple SPST configuration), I see a spark (since the initially uncharged capacitor is virtually a short circuit). So, here is what I think. Some hefty current must initially flow! That causes current to flow at the time of connection, creating Joule hearing in the conductors, plus some radiated energy (like an antenna). And this must be, at least partially, where the missing energy goes. What do you think? I will go on to the next video, hoping to eventually cure my confusion! THANKS FOR THE GREAT VIDEOS!
I have the same problem of missing energy theoretically. The total initial energy stored in the capacitor 1/2 CV(square) is more than the combined energy of both the capacitors. This means some of the energy is being lost?
INDEED, CAPACITOR (& SUPERCAPCITOR) IN THE PAST AND AS OF TODAY HAS BECOME THE MOST SOUGHT AFTER AND RELIABLE ENERGY STORAGE DEVICE IN ELECTRIC POWER AND ELECTRONICS CIRCUIT BOTH IN HV AND LV APPLICATION, RESPECTIVELY WHICH OUTPERFORMS THE BATTERY STORAGE EVEN IN RENEWABLE SOURCE OF ENERGY APPLICATION FOR THE NEEDED INTERFACE WITH THE POWER UTILITY GRID.
sir, i have a question. when we reconnected the c1 to circuit, there is no q on the circuit because q2 has 0c. yet you said while + charges moving the c2 repels + charges. that + charges where did it come ? have circuit charged inital?
Anytime you walk in a circle you end up at the same elevation as you started. Anytime you go around a complete loop in a circuit you end up at the same voltage as you started.
It is a series circuit, but when the 2 capacitors are connected, you can think of it as a parallel circuit in the sense that the voltage across the 2 capacitors must be the same (although one should be negative)
If the two capacitors are connected in such a way that they are in parallel and circuit is completed in battery. What would happen, will the charge still flow from C1 to C2?
The charges would move until the voltages are the same on the two capacitors if they are in parallel, and the charges would move until the charge is the same on the two capacitors if they are in series.
there's a relation between charge q(at general time t)=Q(1-e^(-t/RC)) here Q is charge at steady state R is resistance of wire C is capacitance t is time (general time) if we will equate q with Q and solve the exponential equation we will get the time HOPE IT HELPS !!!
Hello sir. I have a general question. Suppose we have a charged capacitor and a battery along with it in a circuit and . In order to simplify the question we make another battery out of charged capacitor (v=q/c). Then what will be the direction of emf of battery made out of capacitor .
If the capacitor is fully charged and then replaced with a battery, the polarity of the battery will be opposite to the polarity of the initial battery.
@@MichelvanBiezen Thank you Sir. I could not find a video with this same case. A capacitor charging another capacitor, but with a resistor, in series, between them. But anyway, I am very grateful for your content here. Helps a lot.
You can find it here: PHYSICS 48 RC AND RL here: PHYSICS 49 RCL CIRCUITS and there are many videos on that here: ELECTRICAL ENGINEERING 8: RC AND RL CIRCUITS
Steady state is state where the capacitors are full of charges and no more extra charge can be accommodated on em' due to this current flowing through it stops completely eventually bteaking the circuit
any indian student will be laughing on that calculation. here in india we are taught to do calculations very fats and these level of calculations are done sometimes in mind only. No disrespect to you sir, i love your lectures and i come here from time to time . may be you have opted for a slow calculation based on other foreign audiance
Steady state means that enough time has gone by such that the transient state (the time during which the current changes) is done and the current no longer changes. (or changes in a predictable manner in the case of sinusoidal current, or driven by a oscillating source).
This man is a treasure and we must protect him at all costs!
Thank you. Glad you found our videos. 🙂
I always refer to your video when get something confusing and you always help me, thank you sir
Thank you. Glad you like our videos and you find them helpful. 🙂
Sorry for the dumb question sir, but this is just out of my curiosity, why the conservation of charge is applicable to only the inner part between 2 capacitor, not the outer part
Not a dumb question at all. The "why" questions are very important. In the case of a capacitor, the plates always have opposite charges. Therefore the charges on one plate attract the charges on the opposite plate, which causes the charges to migrate to the inner sides of the two plates.
@@MichelvanBiezen hohohohoohoh,, I see alrighty thanks a lot sir....
4:20 The "initial" total charge is ZERO, not q1. It is only q1 if you consider the electrons alone (and they are the only moving charges in a solid). NOTHING forbids them to recombined with an atom momentarily missing an electron (a positive charge relevant to the discussion), or a neutral atom to generate an extra free electron , like they did to generate "q1" in the first place. So, I fail to see the support of the assertion about the "conservation" of the initial amount of negative charge. I agree that we must not destroy electrons, but the "free" ones can recombine and new free ones can be generated.
Keeping the total energy, though, seems more fundamentally grounded, but that leads to a different result.
Since we are dealing with capacitors, the initial charge is as stated in the video. It reprents the extra charge on one side of the capacitor.
Sir what would have happened if we connected them in series instead of parallel arrangement?
They are connected in series.
Hello Professor. Does the statement "V1-V2=0" apply for capacitors in both series and parallel?
Yes I NEED to know the answer to that question too!
That is Kirchhoff's voltage law.. you can look it up.. It says that the algebraic sum of all voltages in a closed loop is 0.. It is applicable for all components regardless of their configuration.. hope this helps :)
this statement is valid everywhere irrespective of configration
You saved me sir..tommorow is my test and this topic was confusing me so much..thankyou so much..
Good luck on your test.
very clean and neat explanation. People like him only should teach physics.
Glad you think so!
Thankssss sir i have stucked in this problem for a month T.T *happy sound*
Welcome 👍
Since the charges in both the capacitors become equal, the energy loss of the charged capacitor is same as energy gained by the uncharged one. Is that right?
That is correct.
However, the total energy decreases from 400uJ to 200uJ!!!
OK, I tried this with different capacitors and initial charges on C1 with identical C2 discharged. I ALWAYS get a SMALLER NUMBER!! One reason I think is, is that when the two caps are connected together (very carefully, with one lead already connected, while the other is connected with a simple SPST configuration), I see a spark (since the initially uncharged capacitor is virtually a short circuit). So, here is what I think. Some hefty current must initially flow! That causes current to flow at the time of connection, creating Joule hearing in the conductors, plus some radiated energy (like an antenna). And this must be, at least partially, where the missing energy goes. What do you think? I will go on to the next video, hoping to eventually cure my confusion!
THANKS FOR THE GREAT VIDEOS!
I have the same problem of missing energy theoretically. The total initial energy stored in the capacitor 1/2 CV(square) is more than the combined energy of both the capacitors. This means some of the energy is being lost?
@@gauravprakash AND, it turns out that 1/2 of the TOTAL ENERGY is always lost!!
INDEED, CAPACITOR (& SUPERCAPCITOR) IN THE PAST AND AS OF TODAY HAS BECOME THE MOST SOUGHT AFTER AND RELIABLE ENERGY STORAGE DEVICE IN ELECTRIC POWER AND ELECTRONICS CIRCUIT BOTH IN HV AND LV APPLICATION, RESPECTIVELY WHICH OUTPERFORMS THE BATTERY STORAGE EVEN IN RENEWABLE SOURCE OF ENERGY APPLICATION FOR THE NEEDED INTERFACE WITH THE POWER UTILITY GRID.
That is defimitely becoming an area of research.
sir, i have a question.
when we reconnected the c1 to circuit, there is no q on the circuit because q2 has 0c. yet you said while + charges moving the c2 repels + charges. that + charges where did it come ? have circuit charged inital?
It might be a mistake. But in general he meant that the charges get shared in between the capacitors according to the capacitance values.
Good illustration. Keep going
why is the voltage must add up to zero in the circuit???
Anytime you walk in a circle you end up at the same elevation as you started. Anytime you go around a complete loop in a circuit you end up at the same voltage as you started.
@@MichelvanBiezen thats kirchoff s loop law and in physics we follow it without any question because its a law
Is it parallel combination because it looks like series
It is a series circuit, but when the 2 capacitors are connected, you can think of it as a parallel circuit in the sense that the voltage across the 2 capacitors must be the same (although one should be negative)
@@MichelvanBiezen thank you sir I got that. The circuit is in series but in a sense the voltage is same so it looks parallel, right?
If the two capacitors are connected in such a way that they are in parallel and circuit is completed in battery. What would happen, will the charge still flow from C1 to C2?
The charges would move until the voltages are the same on the two capacitors if they are in parallel, and the charges would move until the charge is the same on the two capacitors if they are in series.
@@MichelvanBiezen But both are same as u said. If this is parallel then how do we connect in series or vice versa
Hie Michel, How much time it will take to attain steady state i.e, two capacitors gets charged equally after reconnecting the capacitors?
@Pranjil Jain but how much.
there's a relation between charge
q(at general time t)=Q(1-e^(-t/RC))
here Q is charge at steady state
R is resistance of wire
C is capacitance
t is time (general time)
if we will equate q with Q and solve the exponential equation we will get the time
HOPE IT HELPS !!!
Please wich is bigger charging or discharging time ? thank you vety much
They are the same.
thank you for this wonderful explanation, sir! :D
Hello sir. I have a general question. Suppose we have a charged capacitor and a battery along with it in a circuit and . In order to simplify the question we make another battery out of charged capacitor (v=q/c). Then what will be the direction of emf of battery made out of capacitor .
If the capacitor is fully charged and then replaced with a battery, the polarity of the battery will be opposite to the polarity of the initial battery.
This solves my doubt. Thanks a lot sir 🙏✨
Amazing...love from india
Welcome!
what if we have a resistor between the capacitors?
Those are called RC circuits. We have a playlist on that. PHYSICS 48 RC AND RL CIRCUITS PHYSICS 49 RCL CIRCUITS
@@MichelvanBiezen Thank you Sir. I could not find a video with this same case. A capacitor charging another capacitor, but with a resistor, in series, between them.
But anyway,
I am very grateful for your content here. Helps a lot.
What will happen if two capacitor having some charge are connected and then separated
We have a number of examples like that in the playlist.
Michel van Biezen thanks I will see that then..
Sir,do you have any lecture on charging and discharging capacitor and their graph??
Yes, that is extensively covered in the RC circuit playlist and RCL playlist Do you know how to find those from the home page?
@@MichelvanBiezen i am now watching your LR circuit.is it there??
You can find it here: PHYSICS 48 RC AND RL here: PHYSICS 49 RCL CIRCUITS and there are many videos on that here: ELECTRICAL ENGINEERING 8: RC AND RL CIRCUITS
Thanks .IT was really helpful for me. 😄🙏😁
Most welcome 😊
thank u sir, you made it very simple for me.
Glad to hear that
Hello sir, what is steady state
Steady state is state where the capacitors are full of charges and no more extra charge can be accommodated on em' due to this current flowing through it stops completely eventually bteaking the circuit
This was very helpful :)
Glad it was helpful!
Love from VietNam sir
Glad we could help. Welcome to the channel!
Thanks sir your video help me
Welcome. Glad it was helfpul. 🙂
any indian student will be laughing on that calculation. here in india we are taught to do calculations very fats and these level of calculations are done sometimes in mind only. No disrespect to you sir, i love your lectures and i come here from time to time . may be you have opted for a slow calculation based on other foreign audiance
Indeed, the Indian academic standards are very high.
Yes, the calculation has some basic steps. But I wouldn’t laugh though.
Think you Mr
You are welcome.
Love from India🇮🇳
Welcome to the channel!
This video easy to be understood ❤️ #gwalabrand
thank you!!
You are welcome. Glad you found our videos. 🙂
great
🙂
Nice
thaaaaankssss
Hello sir, what is steady state
Steady state means that enough time has gone by such that the transient state (the time during which the current changes) is done and the current no longer changes. (or changes in a predictable manner in the case of sinusoidal current, or driven by a oscillating source).
thank u sir