Q = C . V
I.T = C . V
10 . 0.000027 = C . 3.3
C = 8.8uF
current does not damage the capacitor, the voltage only if exceded will domage the capacitorHowever, I´m not sure if that amount of current could damage capacitor.
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...current does not damage the capacitor, the voltage only if exceded will domage the capacitor...
For a very short time, it should be able to supply, but you cant predict the charging time due to;the question is in other words : can the battery supply more the 10 h current to the capacitor ? in order to fast charging it ?
the question is in other words : can the battery supply more the 10 h current to the capacitor ? in order to fast charging it ?
From the way you present the problem, it seems you like to charge the capacitor manually using a heavy wire to decrease its resistance (though there are already two internal series resistances; one of the battery and the other of the capacitor). If the charging will not be done manually (and heavy wires) then you will need to consider the resistance of the device that will pass the charging current.
Of course you can if there is no maximum limit for the charging time... Otherwise it may be impossible... since the following formula should be satisfied
V = I * T / C
(assuming I, the charging current, is constant. Actually it will decrease exponentially with the time constant RC where R is the sum of all series resistances between the ideal voltage source of the battery 3.3V and the ideal capacitance C of the capacitor).
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