Question based on a video (battery and voltage)

[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.

 

[SammyS]

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 thought.

First, think.

The above formula does specify which is at higher potential.
If $\ Q_1< Q_2 \,,\$ then $\ Q_1- Q_2 \$ and $\ V_x- V_y \$ are negative meaning V_y becomes the V₊.

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.