Find the voltage and energy for each capacitor

[SgtSkittles]

Homework Statement

I am very lost, I tried everything I can do to find the voltage/stored energy for each capacitor, but no luck.

I know that for parallel, V1=V2=V3 etc and stored energy is Q1+Q2 etc. And for series it's the opposite, but I'm getting completely different numbers for both of them.

For the parallel capacitors I'm getting different voltages and for series I'm getting different stored energy. My teacher did give me a hint and said that the voltage for C3 and C5 is 2.493, but I don't understand how he got that and he wouldn't explain it. So I'm hoping someone here can help guide me in the right direction.

Homework Equations

The Attempt at a Solution

I found the total capacitance:

C35 = 3.500

C235 = 1.685

C2345 = 1.207

C12345 = 2.950

C123456 = 1.425

I could only find the voltage and stored energy for C123456 and C6 (by working backwards), however I can't figure out how to calculate the other ones.

I got the information for C123456 by doing the following:

For energy stored I did 1.425 UF * 15v = 21.38UC

And for C6 I got the voltage by doing this:

V = 21.38UC/2.75UF= 7.775v

 

[kuruman]

If the voltage across C6 is 7.775 V, what is it across C1? Remember that V1 + V6 = 15.0 V. Similarly you can sort out the other voltages.

 

[SgtSkittles]

If the voltage across C6 is 7.775 V, what is it across C1? Remember that V1 + V6 = 15.0 V. Similarly you can sort out the other voltages.

Wow, can't believe I didn't think of that. lol. Btw I just tried doing that with the hint my teacher gave, and I'm not getting anything near C3 and C5 = 2.493.

 

[kuruman]

Wow, can't believe I didn't think of that. lol. Btw I just tried doing that with the hint my teacher gave, and I'm not getting anything near C3 and C5 = 2.493.

First you need to find the voltage across the C2345 combo. Use that to find the charge on the C2345 combo. How much charge is on the C35 combo?

 

[SgtSkittles]

First you need to find the voltage across the C2345 combo. Use that to find the charge on the C2345 combo. How much charge is on the C35 combo?

I'm confused, I'll give it a shot though. But, to do that would I use the charge I got for C123456, C6, or C1 to find the voltage for that combo?

 

[kuruman]

How does the voltage across the C2345 combo compare against the voltage across C1? Look at the work you did to find the total equivalent capacitance.

 

[SgtSkittles]

How does the voltage across the C2345 combo compare against the voltage across C1? Look at the work you did to find the total equivalent capacitance.

Well, since C2345 and C1 are parallel, I'm going to say they would compare by the voltage sum of C2345 is equal to the sum of C1. If that's the case, nothing I do, gets 2.493 for C3 and C5. I tried doing 12.651 (charge of C1) divided by 1.207, but just get 10.481 and if I subtract that from 7.229 (C1 Voltage) I get 3.252v

 

[kuruman]

If you know the voltage across C2345, can you find the charge on the C2345 combo? What is it?

 

[SgtSkittles]

If you know the voltage across C2345, can you find the charge on the C2345 combo? What is it?

I got the voltage is equal to 10.48v and the charge is 12.651

 

[kuruman]

Well, since C2345 and C1 are parallel ...

What do capacitors in parallel have that is the same?

 

[SgtSkittles]

What do capacitors in parallel have that is the same?

Voltage!

Then if the voltage is the same, so actually it would be 7.229/1.207 which equals 8.7254 .

Then the charge for C4 would be Charge/4.25 and then I'd do the same for C2.

C4 (charge is 8.7254, voltage 2.053v) and C2 is (8.7254 charge, and 2.685v).

So, now all that would be left is C35. Would that mean that the voltage of those two would be the same as C1?

 

[kuruman]

So, now all that would be left is C35. Would that mean that the voltage of those two would be the same as C1?

Nope. C4, C35 and C6 are in series. What's the charge on C35? Divide that charge by C35 and you have the voltage across C35. See how it works?

 

[SgtSkittles]

Nope. C4, C35 and C6 are in series. What's the charge on C35? Divide that charge by C35 and you have the voltage across C35. See how it works?

Oh. My. God. That makes much more sense. Thank you!