ryan__lai__
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Hey. The answer I got was D as I thought the increase/decrease of PD, current and charge was always exponential. The correct answer however is B. Can anyone explain why?

https://imgur.com/9WvjqAU
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heartx desire
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I suck in this chapter..
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Eimmanuel
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(Original post by ryan__lai__)
Hey. The answer I got was D as I thought the increase/decrease of PD, current and charge was always exponential. The correct answer however is B. Can anyone explain why?

https://imgur.com/9WvjqAU
The charging of a capacitor has an exponential relationship because there is a resistor in series with the capacitor. However, this circuit has the battery in “series” with the capacitor while the resistor is parallel to the capacitor and battery.
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Presto
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But why is it like that?
I don't get how the arrangement affects the reading and I thought that charging/discharging has an exponential relationship only when the current isn't kept constant so when we close the switch wouldn't pd increase exponentially? :/


(Original post by Eimmanuel)
The charging of a capacitor has an exponential relationship because there is a resistor in series with the capacitor. However, this circuit has the battery in “series” with the capacitor while the resistor is parallel to the capacitor and battery.
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Presto
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(Original post by Eimmanuel)
The charging of a capacitor has an exponential relationship because there is a resistor in series with the capacitor. However, this circuit has the battery in “series” with the capacitor while the resistor is parallel to the capacitor and battery.
I think I kinda understand. Please correct me if I'm wrong but the voltage in a parallel circuit is equal in all branches so it wouldn't have to gradually increase and would just spike up... but with that logic, shouldn't it have spiked down when we first closed it?
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Joinedup
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(Original post by Presto)
I think I kinda understand. Please correct me if I'm wrong but the voltage in a parallel circuit is equal in all branches so it wouldn't have to gradually increase and would just spike up... but with that logic, shouldn't it have spiked down when we first closed it?
if you mean when it was first opened... no because the battery is out of circuit when the switch is open. With the switch open it's just a RC circuit with a voltmeter and we need to infer that the capacitor was previously charged up to V from the text.
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Presto
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Yes I meant opened
Oh you're right! So it was in series when the switch was open and parallel when it's closed so if we opened it again it would discharge exponentially in the same way right?
Also just out of curiosity if the switch was open longer, the pd would've eventually decreased to zero right?

(Original post by Joinedup)
if you mean when it was first opened... no because the battery is out of circuit when the switch is open. With the switch open it's just a RC circuit with a voltmeter and we need to infer that the capacitor was previously charged up to V from the text.
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Joinedup
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(Original post by Presto)
Yes I meant opened
Oh you're right! So it was in series when the switch was open and parallel when it's closed so if we opened it again it would discharge exponentially in the same way right?
Also just out of curiosity if the switch was open longer, the pd would've eventually decreased to zero right?
Practically yes: it'll reach a negligible potential in quite a small number of time constants
mathematically no: the exponential decay formula tends towards zero but doesn't reach zero... but I don't think you'll get any questions about the time taken fully discharging a capacitor.
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Presto
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Thank you so much
(Original post by Joinedup)
Practically yes: it'll reach a negligible potential in quite a small number of time constants
mathematically no: the exponential decay formula tends towards zero but doesn't reach zero... but I don't think you'll get any questions about the time taken fully discharging a capacitor.
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