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| Understanding the RC circuit equation |
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| AndrewHodgson:
This is a pretty good explanation: If you want take into account the initial voltage across the capacitor: rearrange that equation so t is the subject t=RC*ln( 1-Vc/Vs) then say you want to know how long it takes you to get from 1V to 4V with a supply voltage of 5V. you say t= RC(1-4/5)-RC*ln(1-1/5) so you're subtracting the time it takes to get to 1V. Make sense? |
| AndrewHodgson:
Also, if you want to take account of the initial capacitor voltage in your derivation you just integrate between t1 and t2, and q1 and q2 |
| Azmarith:
--- Quote from: AndrewHodgson on November 16, 2019, 07:03:07 pm ---This is a pretty good explanation: If you want take into account the initial voltage across the capacitor: rearrange that equation so t is the subject t=RC*ln( 1-Vc/Vs) then say you want to know how long it takes you to get from 1V to 4V with a supply voltage of 5V. you say t= RC(1-4/5)-RC*ln(1-1/5) so you're subtracting the time it takes to get to 1V. Make sense? --- End quote --- --- Quote from: AndrewHodgson on November 16, 2019, 07:44:23 pm ---Also, if you want to take account of the initial capacitor voltage in your derivation you just integrate between t1 and t2, and q1 and q2 --- End quote --- Thank you so much guys! I see where I was going wrong now. I wasn't integrating between t1 and t2, and q1 and q2. I was integrating and just adding on a constant which I assumed was the starting voltage/charge. Thank you all so much for you help ;D |
| bson:
As has been pointed out, just subtract the initial voltage from the supply voltage. Charging a capacitor from 2 to 5V with a 5V supply is exactly the same as charging it from 0 to 3V with a 3V supply. |
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