What does this mean? I don't understand separated again

[flyingpig]

What does this mean? I don't understand "seperated again"

Homework Statement

Charge capacitor C₁ up to some voltage V_1i and then charge another capacitor C₂ up to a different voltage V_2i. Disconnect the capacitors from the supply and connect them in parallel with each other. When the capacitors are again separated what will the final voltage across each capacitor be?

Question on When the capacitors are again separated what will the final voltage across each capacitor be?

What in the world does that mean? Are they asking what the final voltage will be while the capacitors are still hooked up? Or do they mean hook it up and disconnect it again and then measure it (which sounds silly)

 

[gneill]

Silly is good. Capacitors hold a charge even when they're not in a circuit.

Seems clear enough; you charge them to different voltages, connect them, then disconnect them. Then measure the voltages of the disconnected (free standing, isolated) capacitors.

 

[flyingpig]

Why would you disconnect and measure if you are going to measure them anyways?

 

[gneill]

Why would you disconnect and measure if you are going to measure them anyways?

Hey, I said I understood the procedure. I didn't say anything about understanding the motivation!

Perhaps the idea is to get across the point that capacitors can retain a charge even when disconnected from a circuit? Does the lab manual include an Objectives discussion?

 

[flyingpig]

Hey, I said I understood the procedure. I didn't say anything about understanding the motivation!

Perhaps the idea is to get across the point that capacitors can retain a charge even when disconnected from a circuit? Does the lab manual include an Objectives discussion?

No it introduces what a capacitor is and how to discharge it and getting yourself from electrocuted...

 

[flyingpig]

Also my data says

C₁ = 3300uF
C₂ = 2200uF
V_1i = 6.00V
V_2i = 5.00V

V_1f = 5.41V
V_2f = 5.4V

In my notebook I have written the predictions

C₁V_i1 + C₂V_i2 = (C₁+ C₂)V_f

Then I got in my data

V_1f = 5.6V
V_2f= 5.6V

as my predictions. The math leads me to believe that the capacitors were in fact still hooked up

 

[gneill]

No it introduces what a capacitor is and how to discharge it and getting yourself from electrocuted...

So, it's sort of a Darwinian reading comprehension test?

 

[gneill]

Also my data says

C₁ = 3300uF
C₂ = 2200uF
V_1i = 6.00V
V_2i = 5.00V

V_1f = 5.41V
V_2f = 5.4V

In my notebook I have written the predictions

C₁V_i1 + C₂V_i2 = (C₁+ C₂)V_f

Then I got in my data

V_1f = 5.6V
V_2f= 5.6V

as my predictions. The math leads me to believe that the capacitors were in fact still hooked up

So you predicted that the final voltage for both capacitors would be 5.6V, but your measurements in the lab showed that the final voltage was about 5.4V. Is that the issue?

 

[flyingpig]

Uhh somewhat, it's just that my results show what happens when they are connected, not disconnected

 

[gneill]

Uhh somewhat, it's just that my results show what happens when they are connected, not disconnected

Okay, what did you expect to happen when they were disconnected?

 

[flyingpig]

Nothing the charge will still be 6V and 5V respectively. Does the voltage redistribute themselves when it is connected?

 

[gneill]

Nothing the charge will still be 6V and 5V respectively. Does the voltage redistribute themselves when it is connected?

Of course. The voltages (potential differences) are different, so when the capacitors are connected, the charges are going to rearrange themselves accordingly when they find a path to each other. So a current will flow (briefly) while this happens -- current will flow until the potentials equalize.

Remember, components in parallel have the same voltage across them. When they are suddenly "introduced" to each other (by connecting them when they already have differing potential differences across them), then there will be a brief flurry of activity while they level out.

 

[flyingpig]

Of course. The voltages (potential differences) are different, so when the capacitors are connected, the charges are going to rearrange themselves accordingly when they find a path to each other. So a current will flow (briefly) while this happens -- current will flow until the potentials equalize.

Remember, components in parallel have the same voltage across them. When they are suddenly "introduced" to each other (by connecting them when they already have differing potential differences across them), then there will be a brief flurry of activity while they level out.

I am ashamed to overlook something so trivial...

But what about a series? The net voltage will be the sum of the individual voltage of the capacitors, does that mean it is meaningless to connect and disconnect in a series?

 

[gneill]

I am ashamed to overlook something so trivial...

But what about a series? The net voltage will be the sum of the individual voltage of the capacitors, does that mean it is meaningless to connect and disconnect in a series?

Not at all. If you take two previously charged capacitors holding voltages V1 and V2 and connect them in series, the total voltage across the connected pair (i.e. end to end) would be V1 + V2.

If two previously uncharged capacitors (V1 = 0, V2 = 0) are connected in series and then an external supply is connected across the pair, they will both take up the same charge and will have different voltages V1 and V2 which will sum to the external voltage. The two capacitors could then be removed from their connections with the supply and each other and they will retain those (different) voltages.