Capacitor question
A question says 'a 250 micro farad capacitor is fully charged from a 6V battery and then discharged through a 1 kila ohm resistor.
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
Reply 1
Original post
by LisaPereraA question says 'a 250 micro farad capacitor is fully charged from a 6V battery and then discharged through a 1 kila ohm resistor.
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
Is this an A level Question?
Reply 2
12
Original post
by anusanpreaIs this an A level Question?
Yes, my teacher gave it to me. Alevel Physics AQA
Reply 3
Original post
by LisaPereraA question says 'a 250 micro farad capacitor is fully charged from a 6V battery and then discharged through a 1 kila ohm resistor.
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
A way of solving the problem is
Can you see where you go wrong?
Reply 4
Original post
by LisaPereraA question says 'a 250 micro farad capacitor is fully charged from a 6V battery and then discharged through a 1 kila ohm resistor.
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
Calculated the time taken for the charge on the capacitor to fall to 37% of its original value.'
The answer is 0.25 seconds in the book by doing RC, I don't understand this logic, I found RC then multiplied it by 0.37 to get 0.0925s which is wrong, can someone explain the logic behind this please?
yo, RC is the 'time constant' which (for a discharging capacitor) is the time taken for the charge to fall to about 37% of its original value, so you can kind of see why you didn't need to multiply it by 0.37 no?
might make more sense if you know where that comes from so i put a simple lil derivation below:
Reply 5
12
Original post
by mysnowwhitequeenyo, RC is the 'time constant' which (for a discharging capacitor) is the time taken for the charge to fall to about 37% of its original value, so you can kind of see why you didn't need to multiply it by 0.37 no?
might make more sense if you know where that comes from so i put a simple lil derivation below:
might make more sense if you know where that comes from so i put a simple lil derivation below:
Thank you, that really helped
Reply 6
12
Original post
by mysnowwhitequeenyo, RC is the 'time constant' which (for a discharging capacitor) is the time taken for the charge to fall to about 37% of its original value, so you can kind of see why you didn't need to multiply it by 0.37 no?
might make more sense if you know where that comes from so i put a simple lil derivation below:
might make more sense if you know where that comes from so i put a simple lil derivation below:
if I was doing it for charging and it asked for 63% which I just work out T=RC for that one
Reply 7
Original post
by LisaPereraif I was doing it for charging and it asked for 63% which I just work out T=RC for that one
yeah iirc
i think it'd be good if you did what i did above (looking at what happens when t = RC) but with the charging capacitor equation, so you can see why that is
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