This is a slightly theoretical question.
If you charge a capacitor from say a 3V battery you get Q charge in T seconds.
If you double the voltage to 6V you get 2Q charge also in T seconds.
My question is, why doesn't T change? If you think of a capacitor like a room and charge like people, if you want to fit in twice as many people, surely you need more time?
I am thinking of the formula Q1 = Q0e^-(t/RC).
Is there something I am missing?
Best Answer
Current is rate of charge (1 ampere is 1 coulomb/sec). At time zero, the capacitor voltage is zero, so the voltage across your resistance is three volts. The instantaneous current is 3V/R. When you apply 6V (assuming the resistance is equal and ignoring @Whit3rd 's excellent point that adding a second battery will likely increase the resistance), the instantaneous current at time zero would be 6V/R, twice that of the first case. This means that in your analogy, the people would be walking into the room in the first case, and running in the second. In the second case, when the capacitor reaches 3 volts, the voltage across your resistor is (6-3) volts, and the current would be equal to the starting current with the 3 volt supply. At this point the people in the second case would have slowed down and be walking in at the same speed as the people in the first case had started.