Capacitors charged to voltage in parallel

[Ineedahero]

Homework Statement

Suppose you have a capacitor with capacitance C charged to voltage V and a second capacitor with capacitance C charged to voltage 2V. The two capacitors are connected in parallel. What will be the voltage across the two capacitors after being connected in parallel?

Homework Equations

Q = CV

The Attempt at a Solution

These kinds of questions confuse the hell out of me.

What does it have to do with a parallel connection? What's the difference if it was in series?

I know that for a parallel connection, the charge is constant throughout, and for a series connection the voltage is constant throughout...but what does that really mean? How do you apply it?

I really see no possible way to solve this, but at the same time I feel like it's simple.

What are you supposed to do here??

 

[gneill]

Homework Statement

Suppose you have a capacitor with capacitance C charged to voltage V and a second capacitor with capacitance C charged to voltage 2V. The two capacitors are connected in parallel. What will be the voltage across the two capacitors after being connected in parallel?

Homework Equations

Q = CV

The Attempt at a Solution

These kinds of questions confuse the hell out of me.

What does it have to do with a parallel connection? What's the difference if it was in series?

I know that for a parallel connection, the charge is constant throughout, and for a series connection the voltage is constant throughout...but what does that really mean? How do you apply it?

That's exactly backwards: Components in parallel must experience the same potential difference across them (since their connections share the same pair of nodes). Components in series share the same current (since there's nowhere for it to go except follow the single path through them).

Here you've got two charged capacitors with different voltages on them being connected together in parallel. The charges on these capacitors will have to redistribute in order to achieve the same voltage across both (parallel --> same potential difference). Charge is a conserved quantity.

So start by working out the charge on each (symbolically). How will you distribute the total charge so that the capacitors will have the same potential difference?

 

[barryj]

Before we leave this problem, just for fun, calculate the energy stored in the two capacitors before connecting them together and the energy stored in the capacitors after connecting them in parallel. Where did the energy go?

 

[Ineedahero]

That's exactly backwards: Components in parallel must experience the same potential difference across them (since their connections share the same pair of nodes). Components in series share the same current (since there's nowhere for it to go except follow the single path through them).

Here you've got two charged capacitors with different voltages on them being connected together in parallel. The charges on these capacitors will have to redistribute in order to achieve the same voltage across both (parallel --> same potential difference). Charge is a conserved quantity.

So start by working out the charge on each (symbolically). How will you distribute the total charge so that the capacitors will have the same potential difference?

Thanks for the help!

Here's where I'm getting stuck - you say to work out the charge on each. Well, how do I calculate the charge??

I know that q = cv. I also have, essentially, two capacitors connected in series - once I find the equivalent capacitance of the two capacitors connected in parallel.

So that leaves me with two capacitors of 14 connected in series. I know that the charge is constant throughout both of them...but how do I find it??

q=cv = 14 * 12 = 168...but what if the capacitors weren't of equal capacitance? Then that formula wouldn't work...

 

[gneill]

Thanks for the help!

Here's where I'm getting stuck - you say to work out the charge on each. Well, how do I calculate the charge??

I know that q = cv. I also have, essentially, two capacitors connected in series - once I find the equivalent capacitance of the two capacitors connected in parallel.

Don't get ahead of yourself Start by finding the charge on each capacitor before they are connected to each other. What's the total?

So that leaves me with two capacitors of 14 connected in series. I know that the charge is constant throughout both of them...but how do I find it??

q=cv = 14 * 12 = 168...but what if the capacitors weren't of equal capacitance? Then that formula wouldn't work...

I don't know where these numbers are coming from; the problem only uses symbolic values (C,V).

Also, while you can interpret two components connected together to be either in series or in parallel, in this sort of problem you'll find it easier to interpret the connection in parallel fashion. This is because you can make use of the equal potential difference criteria of parallel connections and the equation you already have for the potential across a capacitor (V = q/C).

If the capacitances were not equal then you'd need to write a pair of equations and solve for the individual charges; the total charge being a constant and potentials across them being equal.

 

[Ineedahero]

Don't get ahead of yourself Start by finding the charge on each capacitor before they are connected to each other. What's the total?

Hmm

Q₁ = C₁V₁
Q₂ = C₁*2V₁

Q₁ + Q₂ = 2C₁*3V₁

Is that correct?

 

[gneill]

Right idea, but check your algebra.

 

[Ineedahero]

Of course, what a silly error!

Q = 3C₁V₁

 

[gneill]

Of course, what a silly error!

Q = 3C₁V₁

Okay, that looks better. So now, how will that charge be distributed on the two (equal) capacitors when they are connected in parallel?

 

[Ineedahero]

Well, since the voltage is constant (since they're in parallel), and the capacitances are equal, the charges must also be equal.

So the charge will be cut exactly in half and go to each capacitor?

 

[gneill]

Well, since the voltage is constant (since they're in parallel), and the capacitances are equal, the charges must also be equal.

So the charge will be cut exactly in half and go to each capacitor?

That's correct. So what's the resulting voltage on one of the capacitors?

 

[Ineedahero]

Hmm

Q_each = 1.5*C₁V₁

The capacitance doesn't change, but the voltage does.

Q_each = C₁V₂

1.5V₁ = V₂

So the new voltage is 1.5 * the original voltage?

 

[gneill]

Yup.

 

[Ineedahero]

Ok!

Thanks a lot for the help!