# PD/Current of Capacitors

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#1
http://tinypic.com/r/nn1xlj/8

Hi, how do i know how to distribute the PD and Current across each of these capacitors?

I know how to find the total capacitance of a series and parallel circuit, but i'm not sure about the order of which i should apply the formulas.
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6 years ago
#2
(Original post by minnigayuen)
http://tinypic.com/r/nn1xlj/8

Hi, how do i know how to distribute the PD and Current across each of these capacitors?

I know how to find the total capacitance of a series and parallel circuit, but i'm not sure about the order of which i should apply the formulas.
Look at the circuit closely. How is the 300uF capacitor connected the supply?

Will the charge on the 300uF capacitor be affected in any way by the other capacitors?

Now turn your attention to the series capacitors:

Calculate the equivalent capacitance for the series combination and using the supply voltage you can calculate the total charge on that series combination.

What can you say about how the total series charge is shared between those two series capacitors?

HINT: look at the central plates of the two capacitors and ask where the charge for those isolated plates is going to come from?
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#3
(Original post by uberteknik)
Look at the circuit closely. How is the 300uF capacitor connected the supply?

Will the charge on the 300uF capacitor be affected in any way by the other capacitors?

Now turn your attention to the series capacitors:

Calculate the equivalent capacitance for the series combination and using the supply voltage you can calculate the total charge on that series combination.

What can you say about how the total series charge is shared between those two series capacitors?

HINT: look at the central plates of the two capacitors and ask where the charge for those isolated plates is going to come from?
I think that because the current is shared between branches in a parallel circuit, that the charges on each branch will affect each other.

The equivalent capacitance of the series branch is 112.5F. Thus, i think that the charge across that branch and each capacitor will be Q= VC= 6x112.5= 675C.

That's what i've got so far . I know that the voltage for the 2nd branch will be the same for the 1st branch, could you help a bit further please?
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6 years ago
#4
(Original post by minnigayuen)
I think that because the current is shared between branches in a parallel circuit, that the charges on each branch will affect each other.
The supply can be thought of as an infinite source of current and charge. Since the 300F capacitor is placed directly across the supply and is in parallel with the other capacitors, then the pd across the 300F capacitor will also be that of the supply voltage.

So both the charging current and the final charge on the 300uF capacitor will NOT be affected by the other two series capacitors.

So the charge on the 300uF capacitor is Q=CV = 300x6 = 1800C

(Original post by minnigayuen)
The equivalent capacitance of the series branch is 112.5F.
Correct

(Original post by minnigayuen)
Thus, i think that the charge across that branch and each capacitor will be Q= VC= 6x112.5= 675C.
The total charge of 675C is shared between the two series capacitors.

So to get the pd developed across each capacitor, go back to the Q=CV equation and rearrange to make V the subject.

Then substitute for the charge just calculated (675C) and the original value of each individual series capacitor.

To check you have the right answers, add up the pd's and they should equal the supply voltage.
1
#5
(Original post by uberteknik)
The supply can be thought of as an infinite source of current and charge. Since the 300F capacitor is placed directly across the supply and is in parallel with the other capacitors, then the pd across the 300F capacitor will also be that of the supply voltage.

So both the charging current and the final charge on the 300uF capacitor will NOT be affected by the other two series capacitors.

So the charge on the 300uF capacitor is Q=CV = 300x6 = 1800C

Correct

The total charge of 675C is shared between the two series capacitors.

So to get the pd developed across each capacitor, go back to the Q=CV equation and rearrange to make V the subject.

Then substitute for the charge just calculated (675C) and the original value of each individual series capacitor.

To check you have the right answers, add up the pd's and they should equal the supply voltage.
Thank you so much kind sir! 0
6 years ago
#6
(Original post by minnigayuen)
Thank you so much kind sir! You are most welcome.

Just as a matter of completeness, are the values of capacitance in your diagram correct?

1F is a huge capacitance which would normally be stated in F (microFarads = 10-6F i.e. millionth parts of a Farad.)
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#7
(Original post by uberteknik)
You are most welcome.

Just as a matter of completeness, are the values of capacitance in your diagram correct?

1F is a huge capacitance which would normally be stated in F (microFarads = 10-6F i.e. millionth parts of a Farad.)
Oh yeah, no sorry, they would be microfarads but i didn't have a source for this question - i produced it myself and was just saving time!
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