But if we take initial V to be 5, and final V to be 0.7, then isn't the time delay going to be less than 1 minute? because by the time it gets to 0.7V it is already below the threshold amount(4.3)required for the circuit to be active. Doesn't the capacitance need to have at least 4.3V for the circuit to be active? meaning we must take 4.3 to be final V? Also, how do other members in this forum add special mathematical symbols?
(Original post by 99llewellyn99)
in this question the capacitor has to be charged up to at least 4.3V before the circuit becomes active this has to be achieved in a time of 60 seconds. we only have learnt the formula for a capacitor being discharged so we can think of the cell as being discharged instead so for a 5V cell that has dissipated 4.3V to a capacitor only 0.7V will be left in the cell so we can use the formula
now you can find R you should get a value of 1.39x104
which is approximately 13.9 kilo-ohms
Last edited by osake; 02-06-2012 at 16:00.