# Capacitor circuit

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Question as above:

(i)

(ii)

(iii)

(iv) When the capacitor is fully charged. It will charge faster than the capacitor will discharge.

I'm confused as to what the word 'final' means? Surely, finally the charges, energy and potential difference will all be 0, because energy will be lost somewhere in the system.

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#2

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(Original post by

Remember when finding the total capacitance of a group of capacitors, the rules are opposite to that of resistors.

**Phichi**)Remember when finding the total capacitance of a group of capacitors, the rules are opposite to that of resistors.

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#5

(Original post by

Any help?

**halpme**)Any help?

v) 'final' You need to work out the charge on each capacitor when the current stops flowing. i.e. when steady state conditions are achieved.

V = Q/C

and since they are in parallel, the p.d. across each capacitor must be equal. i.e. the voltage 'pressure' on the two capacitors is equalised.

The rest should follow on form that.

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#6

(Original post by

Any help?

**halpme**)Any help?

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(Original post by

How far have you got with this since my last post?

**uberteknik**)How far have you got with this since my last post?

With regard to iv), I was thinking that if one of the capacitors was fully charged, it couldn't take anymore charge, and would therefore stop discharging (somehow?!), obviously incorrect. If both of them were fully charged simultaneously though, no charge would flow because they'd both be at their respective maximum charges?

In my mind, I can't see why there'd be a difference between the answers from .

I assume the circuit now has an overall effective capacitance, and therefore the 'final' potential difference between the plates of the capacitor will change?

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#8

(Original post by

Not very far, I have to say, I moved onto some other questions.

With regard to iv), I was thinking that if one of the capacitors was fully charged, it couldn't take anymore charge, and would therefore stop discharging (somehow?!), obviously incorrect. If both of them were fully charged simultaneously though, no charge would flow because they'd both be at their respective maximum charges?

In my mind, I can't see why there'd be a difference between the answers from .

I assume the circuit now has an overall effective capacitance, and therefore the 'final' potential difference between the plates of the capacitor will change?

**halpme**)Not very far, I have to say, I moved onto some other questions.

With regard to iv), I was thinking that if one of the capacitors was fully charged, it couldn't take anymore charge, and would therefore stop discharging (somehow?!), obviously incorrect. If both of them were fully charged simultaneously though, no charge would flow because they'd both be at their respective maximum charges?

In my mind, I can't see why there'd be a difference between the answers from .

I assume the circuit now has an overall effective capacitance, and therefore the 'final' potential difference between the plates of the capacitor will change?

Now think about e.m.f. and voltage. This is a measure of the force available to do work.

The 8uF has 120V potential to do work and the 4uF capacitor has no potential. When the switch is closed at t=0, a circuit is completed and the 8uF cap' has the energy needed to force current around the circuit (and hence build) charge onto the 4uF cap'.

But as that charge starts to flow and build on the 4uF cap, it gets it's charge from the 8uF cap'. hence the p.d. developed on the 4uF cap increases while the charge and p.d. on the 8uF cap' falls.

Current will stop flowing when the voltage on both capacitors is in equilibrium. i.e. the p.d. across both is the same.

We know the relationship V = Q/C, and that Q

_{total}= Q

_{8uF}+ Q

_{4uF}, therefore you can set up a simple equilibrium equation relating the ratio of charges to capacitance. i.e. the total charge calculated in part i) must be shared between the capacitors (energy conservation) and also knowing that the voltage on both capacitors must be the same when equilibrium is achieved.

Can you now see how to do this?

Spoiler:

Show

After the switch is closed, the total charge held on the 8uF is then shared with the other 4uF capacitor. However, the total capacitance has increased which you can calculate since they are now in parallel.

So you can now calculate the voltage on the parallel capacitors because you already know the total charge.

And when you know the p.d. across the pair, you can calculate the individual charges because the capacitances are different.

So you can now calculate the voltage on the parallel capacitors because you already know the total charge.

And when you know the p.d. across the pair, you can calculate the individual charges because the capacitances are different.

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