How in the world did I get the right answer? I am scared

[flyingpig]

Homework Statement

Again this is another lab problem, but this time it is even more scary because my measurements and calculation match, which shouldn't because my calculation is wrong.

Charge a capacitor C₁ 3300uF to some voltage V_1i, disconnect it from the supply and connect it in parallel with the second capacitor C₂, which is initially uncharged. When the capacitors are again separated what will the final voltage across each capacitor be?

Data Collected

C_1i = 3300uF
C_2i = 2200uF
V_1i = 6.00V
V_2i = 0

V_1f = 3.5V
V_2f = 3.2V

Prediction

In my notebook I had

(I) C_1 = Q \Delta V_1

(II) C_2 = Q \Delta V_2

Divided (I) by (II)

\frac{C_1}{C_2} = \frac{\Delta V_1}{\Delta V_2}

\frac{C_1}{C_2} = \frac{V_{1f} - 6.00V}{V_{2f} - 0}

Now somehow I got from that equation to

\frac{C_1 V_{1i}}{C_1 + C_2} = V_{2f}

Which looks like a conservation of charge equation

Then the results I got were

V_1f = 3.6V

V_2f = 3.6V

The Headache

Someone I assumed that the charge were the same in a parallel circuit and I did the math. Now I don't know what happened, but I got 3.5V and 3.2V, which is really close to my measurements and now I am scared. WHAT IN THE WORLD HAPPENED?

I know this is a very bad question asking "figure it out for me", but I am at my limits here

 

[flyingpig]

Wait I just got it, I did it right never mind. In case you are wondering I did...

Q_1 + Q_2 = Q_{net}

\Sum \Delta Q = 0

C_1 V_{1i} + C_2 V_{2i} = Q_{net} = C_1 V_{1f} + C_2 V_{2f}

But V is the same everywhere in this circuit and V₂ was 0

\frac{C_1 V_{1i}}{C_1 + C_2} = V_{f}

In my book I wrote V₂ instead of V_f...that's why I was confused.

I just killed a thread because of one oversight...oops

 

[gneill]

Increasingly, I can't tell what is lab measured data and what is your calculated predicted values in your posts. Above you put 3.5V and 3.2V as "collected data", and later you say that these values were really close to your measurements. So what are your measurements, if not "collected data"? Are the measurement values the 3.6V values that you called "the results I got", which follow after all your calculations?

 

[flyingpig]

I just realize something, does the circuit look like this?

http://img714.imageshack.us/img714/9694/parallel.th.png

Uploaded with ImageShack.us

Is it implied a potential difference was attached to the circuit? Does it even need a power supply?

 

[gneill]

That diagram does not appear to correspond to any phase of the experiment that you've described so far.

 

[flyingpig]

What do you mean...?

 

[gneill]

At what point in time, according to the description you've given for the procedure, were both capacitors connected in parallel with a voltage supply?

 

[flyingpig]

At what point in time, according to the description you've given for the procedure, were both capacitors connected in parallel with a voltage supply?

That's what I am worried about...

How about this one? I wipe out the voltage source.

http://img832.imageshack.us/img832/8879/parallel2.th.png

Uploaded with ImageShack.us

 

[gneill]

Yes, that could depict when the two capacitors were connected in parallel. That would be when some charge from C1 moved to C2 (because C2 started with a lower potential difference -- zero actually).

 

[flyingpig]

But isn't that also series? I am very confused now

 

[gneill]

But isn't that also series? I am very confused now

The two ends of each device are connected. They share a potential difference. They are in parallel.

 

[flyingpig]

The two ends of each device are connected. They share a potential difference. They are in parallel.

Does that mean this is in series? Isn't it just how you look at the circuit?

http://img148.imageshack.us/img148/4425/series2.th.png

Uploaded with ImageShack.us

 

[gneill]

They are also in parallel. Both ends of each component are connected to each other. They share a common potential difference.

Note that if you introduce another component in the chain, then they would be in series; You could identify distinct potential differences for each component.

 

[flyingpig]

Like this?

http://img651.imageshack.us/img651/3889/daso.th.png

Uploaded with ImageShack.us

 

[flyingpig]

Also there is a followup question

Use the results to calculate the energy stored in Capacitor C₁ after it has been charged up to some voltage V₁ and then calculate the total energy stored by C₁ and C₂ after they have been connected together and separated. Is the energy stored initially the same as the energy stored after? If not, why?

Calculations

\frac{1}{2} C_1 \Delta V_{1i}^2 = \frac{1}{2} (3300\mu F) (6V)^2 = 0.0594J

\sum U_f = \frac{1}{2} C_1 \Delta V_{1f}^2 + \frac{1}{2} C_2 \Delta V_{2f}^2 = \frac{1}{2} (3300\mu F)(3.5V)^2 + \frac{1}{2}(2200\mu F)(3.2)^2 = 0.0315J

So the excess energy is 0.0594J - 0.0315J = 0.0279J

Now the energy is different, but what is this excess energy? I just connected the circuit and I disconnected again, does the wire eat up some of the energy when the potential difference is being distributed?

 

[gneill]

So 3.5V and 3.2V *are* your lab measured values? You never did answer my question as to what are your measured and what are your predicted values.

Yes, energy will be lost when the charges get rearranged. It disappears as heat in the wires due to the current, and as electromagnetic energy being "broadcast" from the wiring.

 

[flyingpig]

Yes 3.5V and 3.2V are my measured values. My predicted ones were 3.6V

So is the excess joules the internal thermal energy or the work required to rearrange the charges?

 

[gneill]

Yes 3.5V and 3.2V are my measured values. My predicted ones were 3.6V

So is the excess joules the internal thermal energy or the work required to rearrange the charges?

The "excess" (actually a deficit) energy is energy that is lost to the environment in one way or another when the charges "fall" into their new configuration. It's kind of like rolling a rock downhill into a valley; some energy gets lost to friction, sound, crushing things, ...

In a circuit, the biggest culprit is usually resistive losses.

 

[flyingpig]

Isn't that internal thermal energy...?

 

[gneill]

Isn't that internal thermal energy...?

It doesn't stay internal. Components radiate their generated heat to the environment.

 

[flyingpig]

What is the correct wording then?

"The excess energy is the heat transferred from the internal thermal energy to the environment"

Does that work?

 

[gneill]

What is the correct wording then?

"The excess energy is the heat transferred from the internal thermal energy to the environment"

Does that work?

What do you mean by "work"? Stuff gets hot. Stuff cools down.

 

[flyingpig]

In proper English, how do I answer it correctly is what I am asking. In your analogy with a rock rolling downhill, what are the 'sound', 'friction', and 'crushing things' in my circuit?

 

[gneill]

In proper English, how do I answer it correctly is what I am asking. In your analogy with a rock rolling downhill, what are the 'sound', 'friction', and 'crushing things' in my circuit?

"Resistive and electromagnetic emission losses in the wiring and components" should do nicely.

 

[flyingpig]

I am a freshman...my TA is going to raise his eyebrows if he sees such fancy words.

 

[gneill]

I am a freshman...my TA is going to raise his eyebrows if he sees such fancy words.

Ha!

How about "energy losses due to wire resistance and non-ideal components".

 

[flyingpig]

So much better lol, but isn't that what I have been saying?

 

[flyingpig]

I have a question, since my \Delta E is negative, which make sense right?

I had 3.15 x 10^-2 J - 5.94 x 10^-2 J = 2.79 x 10^-2J = -\Delta TE

Does that make better sense?

 

[gneill]

As long as you explain what the quantity is that you're specifying, it's all good.

Personally I would simply say that ΔE = -2.79 x 10^-2 J, where ΔE is the change in stored potential energy of the system (PE_final - PE_initial).

 

[flyingpig]

But it is also equal to the change in thermal energy yush?