Question based on a video (battery and voltage)

[gracy]

in all three cases there is no battery or voltage source,right?Because I can't see any battery.But there is voltage ,then without battery how is potential difference created?I think potential difference is generated due to charged capacitors not by battery or any other voltage source.

 

[cnh1995]

I think potential difference is generated due to charged capacitors not by battery or any other voltage source.

Yes.In the circuit, the p.d. is due to the voltage across the charged capacitor. But for charging the capacitor before connecting it in the circuit, a battery might have been used.

 

[gracy]

Yes.In the circuit, the pd is due to the voltage across the charged capacitor

In all three cases?

 

[cnh1995]

In all three cases?

yes.

 

[gracy]

Thanks @cnh1995

 

[cnh1995]

You're welcome!

 

[gracy]

In the above video for case 1 $V_f$ was potential across equivalent capacitance of $C_1$ and $C_2$ why he used the same $V_f$ for individual capacitors also?If there would have been any battery present like this

then in this case I understand how potential difference across $Ceq$ and individual capacitors is same.But we know it does not have any battery then how potential across equivalent capacitance of $C_1$ and $C_2$ and individual capacitors is same?

 

[cnh1995]

In the above video for case 1 $V_f$ was potential across equivalent capacitance of $C_1$ and $C_2$ why he used the same $V_f$ for individual capacitors also?If there would have been any battery present like this

View attachment 95582

then in this case I understand how potential difference across $Ceq$ and individual capacitors is same.But we know it does not have any battery then how potential across equivalent capacitance of $C_1$ and $C_2$ and individual capacitors is same?

The current will stop only when the voltages of C1 and C2 will be equal. Also, equivalent capacitane is connected between same two points as are C1 and C2.

 

[gracy]

Is potential difference across equivalent capacitor always same as potential difference across it's individual component capacitors?

 

[cnh1995]

Is potential difference across equivalent capacitor always same as potential difference across it's individual component capacitors?

For parallel capacitors, yes.

 

[gracy]

For parallel capacitors, yes.

Even when the potential difference is not due to /created by battery?

 

[sophiecentaur]

in all three cases there is no battery or voltage source,right?Because I can't see any battery.But there is voltage ,then without battery how is potential difference created?I think potential difference is generated due to charged capacitors not by battery or any other voltage source.

Really , Gracy; you expect us to trawl through 8+ minutes of that really boring presentation, which you found confusing. Couldn't you just be kind to us and ask the actual question that the video raised in your mind? The Web is full of that sort of badly presented stuff that, presumably, some poor students are expected to get their education from. From your past posts, I can see you are not clueless and that you have a fair grasp of this subject. I am not just being grumpy when I ask you to ask specific questions in your own words. As my hero Old Jim often remarks, "a question well asked, is half answered". He's sooooo right. This stuff is well within your capabilities, I'm sure.

 

[cnh1995]

Even when the potential difference is not due to /created by battery?

Yes. Because there is no current in the circuit once the capacitors reach steady state. Current ceases to flow only when the capacitors have equal voltage.

 

[cnh1995]

As my hero Old Jim often remarks, "a question well asked, is half answered". He's sooooo right.

True! And Jim is one of the best science advisors I have known here on PF!

 

[gracy]

Gracy; you expect us to trawl through 8+ minutes of that really boring presentation, which you found confusing. Couldn't you just be kind to us and ask the actual question that the video raised in your mind? The Web is full of that sort of badly presented stuff that, presumably, some poor students are expected to get their education from

Well ,I find that video both interesting as well as useful and I am not poor.To each his own.

I wonder how you found out that it is for poor students.It is for educational purpose and we should not classify it as you did .Education is same for all poor or rich.

 

[jbriggs444]

Well ,I find that video both interesting as well as useful and I am not poor.To each his own.

I wonder how you found out that it is for poor students.It is for educational purpose and we should not classify it as you did .Education is same for all poor or rich.

In the vernacular, the "poor whoever" phrase does not indicate that the person is poor financially, but that the situation being described is one which does not act to his or her benefit.

e.g. "oh, those poor people in the oncoming lanes of traffic heading toward that backup that we just passed" applies equally well when those poor people are driving rusted out beaters or new model BMW's.

 

[gracy]

In the vernacular, the "poor whoever" phrase does not indicate that the person is poor financially, but that the situation being described is one which does not act to his or her benefit.

Obviously I know that.But the way he has written ,it looks like he did mean poor( financially).

 

[sophiecentaur]

But the way he has written ,it looks like he did mean poor( financially).

I am so sorry it came across that way. I meant 'poor' in the sense that the students would have been suffering with such a boring presentation.
Our common English language is full of pitfalls.

 

[cnh1995]

students would have been suffering with such a boring presentation.

Well, I liked that post because of its last few lines and Jim's quote. But I disagree with what you said in the first part. Maybe you find his Indian accent boring but I don't think all the students are getting bored with such presentations. You are an old, experienced man. You might have gone through a different learning environment and hence, you might be finding these videos boring, but that doesn't mean all the students should find them boring. "Boring" and "bad presentation" are your personal opinions but with all due respect, I do not think you should express them that way. Some videos like that have been very helpful for me and many of my fellow students. So, I personally think that remark is somewhat rude.

 

[jbriggs444]

Obviously I know that.

It may have been obvious to you. You may have expected it to be obvious to others. However, if it was obvious to me, I would not have attempted an explanation.

It is often useful when interacting over the Internet to assume good intent -- to grow a thick skin and allow imagined slights, insults or misrepresentations to go unchallenged.

 

[sophiecentaur]

Maybe you find his Indian accent boring but I don't think all the students are getting bored with such presentations.

No problem with the accent (My feed was with a very 'English' accent). I just have a problem with presentations that would be better in written form. I am of the opinion that poorly handwritten stuff on a blackboard, shown of a small screen and with limited resolution is nothing like as useful as a well prepared and written passage in a textbook. I guess it's something to do with the culture of needing to see stuff on a screen to get peoples' attention, never mind the quality.
I have been given loads of formal advice about presentation in teaching and delivering technical stuff to colleagues and the way not to do it is to read out exactly what is being written on a board or in a Powerpoint presentation. Perhaps it's too much to ask for some enthusiasm from a lecturer??
I could watch Feynman all day because he does it right.
I am sorry if you found my remark rude but, if people want to get useful input on PF then it would help if the meat of the questions is readily accessible. I think you have to admit that there is an awful lot of very turgid stuff on the web that's just too much like hard work to follow.

 

[cnh1995]

I could watch Feynman all day because he does it right.

That's what I meant by "different learning environment". Feynman's lectures are one of the very best, no doubt about that but they are not shown in every institute and not everyone is Feynman. I also totally agree with your opinion about a Powerpoint presentation but the video is not so bad that one would not understand a thing or worse, the concepts are being taught all wrong. Everyone has his own way, with some flaws. But still the videos are helpful. I agree with your points about a presentation and enthusiasm, but I would not compare him with Feynman or Walter Lewin. The concepts taught by him are perfect and hence the videos are helpful for many.

 

[gracy]

One more question from the same video

From time 5:11 i.e in third case

From time 7:22 to 7:32 he says if we take $Q_1$ >$Q_2$ then $V_x$ > $V_y$

I want to ask why he did not assume $q_{1f}$ > $q_{2f}$ rather than $Q_1$ >$Q_2$? Had he assumed $q_{1f}$ > $q_{2f}$ the result would have been different or the same $V_x$ > $V_y$?

 

[SammyS]

One more question from the same video

From time 5:11 i.e in third case

From time 7:22 to 7:32 he says if we take $Q_1$ >$Q_2$ then $V_x$ > $V_y$

I want to ask why he did not assume $q_{1f}$ > $q_{2f}$ rather than $Q_1$ >$Q_2$? Had he assumed $q_{1f}$ > $q_{2f}$ the result would have been different or the same $V_x$ > $V_y$?

Think about it for a moment. Q₁ and Q₂ are the initial charge values. It only makes sense to begin at the beginning.

But there's more to say regarding your question. Notice that Mr. Arora says nothing to compare $\ q_{1f} \$ and $\ q_{2f} \ .\$ He only gives a result for their sum.

Whether .$\ q_{1f}>q_{2f} \$ or $\ q_{1f}<q_{2f} \$ or even $\ q_{1f}=q_{2f} \$, all of that depends upon the relative capacitances, C₁ and C₂ .

 

[gracy]

Whether . q1f>q2f q1f<q_{2f} or q_{1f}=q_{2f} \ , all depends upon the relative capacitances, C1 and C2 .

But I am assuming that $q_1f$ >$q_2f$ ,if that's the case will $V_x$>$V_y$? By $V_x$ and $V_y$ I mean final potential at x and final potential at y respectively. I think I should write

But I am assuming that $q_1f$ >$q_2f$ ,if that's the case will $V_{xf}$>$V_{yf}$?

 

[SammyS]

But I am assuming that $q_1f$ >$q_2f$ ,if that's the case will $V_x$>$V_y$ ?

Those two inequalities have absolutely nothing to do with each other.

What he is assuming is that the net charge in the left hand side of the circuit is positive (so the right hand side is negative). That gives the relative potential as he states.

 

[gracy]

I have edited my post .

 

[gracy]

What he is assuming is that the net charge in the left hand side of the circuit is positive (so the right hand side is negative). That gives the relative potential.

I got your point. Thanks

 

[cnh1995]

Not true.

Hence deleted..

 

[gracy]

There will be potential different between points x and y given by $\frac{Q1-Q2}{C1+C2}$
Right?

 

[SammyS]

Because there will be potential different between points x and y given by $\frac{Q1-Q2}{C1+C2}$
Right?

Right.

and Q₁ > Q₂

 

[cnh1995]

Actually, I hadn't watched the 3rd case earlier. I thought Vx and Vy would be voltages across the capacitors. Then I watched the video and saw that they are two points. What I actually wanted to say is Vc1 has to be equal to Vc2. Sorry!

 

[gracy]

From time 7:22 to 7:32 he says if we take $Q_1$>$Q_2$ then $V_x$ >$V_y$

This is for initial condition, right? Because finally it does not matter whether $Q_1$ is greater or $Q_2$ , final potential at x will always be greater than final potential at y. Right?

 

[cnh1995]

This is for initial condition, right? Because finally it does not matter whether $Q_1$ is greater or $Q_2$ , final potential at x will always be greater than final potential at y. Right?

If Q1 is greater, voltage polarity of C1 will not change in the end.

 

[gracy]

I think so. If Q1 is greater, voltage polarity of C1 will not change in the end.

And even if Q1 isn't greater final potential at x will still be greater than final potential at y. Right?

 

[gracy]

What are these pictures about?

 

[cnh1995]

And even if Q1 isn't greater final potential at x will still be greater than final potential at y. Right?

I don't think so. If Q1 is greater, then only Vxf>Vyf. You can see it here.

 

[gracy]

$q_{1f}$ + $q_{2f}$ =$Q_{1f}$ - $Q_{2f}$

If $Q_{1f}$ < $Q_{2f}$

It means $q_{1f}$ < $q_{2f}$

But it does not tell much about polarity of $q_{1f}$.

 

[cnh1995]

$q_{1f}$ + $q_{2f}$ =$Q_{1f}$ - $Q_{2f}$

If $Q_{1f}$ < $Q_{2f}$

It means $q_{1f}$ < $q_{2f}$

But it does not tell much about polarity of $q_{1f}$.

See #37. The polarity of 10uF has changed. If Q1 is more, it will retain its polarity.

 

[cnh1995]

If Q1 is greater, Vxf>Vyf. In #37, Q1(charge on 10uF)is less than Q2(charge on 5uF). Hence, polarity of C1(10uF) changed, making Vxf<Vyf.

 

[gracy]

I am unable to comprehend those pictures in post #37.

 

[gracy]

eater, Vxf>Vyf. In #37, Q1(charge on 10uF)is less than Q2(charge on 5uF). Hence, polarity of C1(10uF) changed, making Vxf<Vyf.

How one Can prove it ? I mean besides experimentally.

 

[cnh1995]

There will be potential difference between points x and y given by (Q1−Q2)/(C1+C2)

If Q1<Q2, Vxy will be negative i.e Vx<Vy.

 

[cnh1995]

I am unable to comprehend those pictures in post #37.

It is a simulation result. It is an example of the case 3 in that video. In that circuit, Q1(charge on 10uF) is less than Q2(charge on 5uF) but initially Vx>Vy. When the middle switch is closed, it becomes the case 3. You can see the polarity of the 10uF reversed in the steady state, making Vx<Vy.

 

[gracy]

Can you give me link to that simulation site?

 

[cnh1995]

Can you give me link to that simulation site?

It is an android app on my phone.

 

[SammyS]

gracy,

When you said the following (way) back in Post #30,

There will be potential different between points x and y given by $\frac{Q1-Q2}{C1+C2}$
Right?

I thought it was clear to you as to why V_x > V_y . $\$ After all, V_x − V_y = (Q₁ − Q₂)/(C₁ + C₂) .

 

[gracy]

I thought this formula is for potential difference and it is not particularly $V_x$ - $V_y$ rather $V_+$ - $V_-$. I know it's bit odd but that's what I thought.

 

[SammyS]

I thought this formula is for potential difference and it is not particularly $V_x$ - $V_y$ rather $V_+$ - $V_-$. I know it's bit odd but that's what I thought.

They are the very same thing in this case.

 

[gracy]

When I thought it is $V_+$ - $V_-$

By $V_+$ I mean potential at either x or y whichever is at greater (positive potential ).

($Q_1$ - $Q_2$)/($C_1$ + $C_2$ )This formula just gives potential difference and does not specify anything about which point is at greater potential.This is what I thought.

I wrote there will be potential different between points x and y given by ($Q_1$ - $Q_2$)/($C_1$ + $C_2$ ) because at that time I knew x is at greater potential.