jonnypdot
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When we have 2 different capacitors discharging is there a different approach when solving problems. Also can a capacitor discharge through a second capacitor which discharges through a resistor?
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

You can also find the Exam Thread list for A-levels here and GCSE here. :dumbells:


Just quoting in Puddles the Monkey so she can move the thread if needed
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uberteknik
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(Original post by jonnypdot)
When we have 2 different capacitors discharging is there a different approach when solving problems. Also can a capacitor discharge through a second capacitor which discharges through a resistor?
You need to post a circuit / diagram to illustrate exactly what you mean before an answer can be given. The terminology you have used is rather ambiguous without it. Thanks.
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jonnypdot
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(Original post by uberteknik)
You need to post a circuit / diagram to illustrate exactly what you mean before an answer can be given. The terminology you have used is rather ambiguous without it. Thanks.
standard circuit with 2 capacitors in series with a battery and switch, a resistor is parallel to both capacitors, when capacitors are fully charged and switch is open is there any difference in the discharge?
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uberteknik
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(Original post by jonnypdot)
standard circuit with 2 capacitors in series with a battery and switch, a resistor is parallel to both capacitors, when capacitors are fully charged and switch is open is there any difference in the discharge?
The discharge rate behaviour will still follow the same exponential decay. But because the two capacitors are connected in series, the series capacitance of the combination must be used for both charge and discharge calculations:

\frac{1}{C_{series}} = \frac{1}{C_1} \ + \ \frac{1}{C_2}

Think about where the charge is stored on two capacitors connected in series?

Specifically look at how the capacitor plates are connected between the two capacitors:

These plates are isolated. Charge storage is effectively limited to the size of the isolated plates. Which means the isolated plates can only move a limited amount of charge determined by the smallest isolated plate.

i.e. the total charge stored will be governed by the smallest capacitor in series and it's this capacitor that determines the p.d. developed across the other larger capacitors in the series chain.
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jonnypdot
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(Original post by uberteknik)
The discharge rate behaviour will still follow the same exponential decay. But because the two capacitors are connected in series, the series capacitance of the combination must be used for both charge and discharge calculations:

\frac{1}{C_{series}} = \frac{1}{C_1} \ + \ \frac{1}{C_2}

Think about where the charge is stored on two capacitors connected in series?

Specifically look at how the capacitor plates are connected between the two capacitors:

These plates are isolated. Charge storage is effectively limited to the size of the isolated plates. Which means the isolated plates can only move a limited amount of charge determined by the smallest isolated plate.

i.e. the total charge stored will be governed by the smallest capacitor in series and it's this capacitor that determines the p.d. developed across the other larger capacitors in the series chain.
so
Q=Qoe^t/tau Qo=Ctotal x Emf
to find pd at any given time find Q at that time, Q/C1 =Pd around 1, Q/C2 =pd around 2, emf-pd1-pd1= pd around resistor
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jonnypdot
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can i you also clear this up for me,
we have 2 positive charges which are not equal, and are separated by distance d, the electric field strength at the mid point of their separation will be just the sum of their electric fields at a distance d/2?

If charges were now both negative i would have to take into account direction of the E fields, if they were unlike charges will i also add the E fields?
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