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Apr3-12, 05:43 PM
That's nearly right. Certainly the p.d. across the resistor falls as the p.d. across the capacitor rises when it's charging. But what about the case where the capacitor is discharging? In that case the battery is removed and replaced with a piece of wire (the circuit is different). The capacitor starts out with some initial p.d. across it, and the resistor is connected directly across the capacitor. So the resistor
with the same p.d. as the capacitor...
hold on...when its charging the resistor is not involved
it is only involved during discharging
surely your approach is wrong?