Topic 17. Charging Capacitors
In the investigation, the flow is too fast to see exactly what happens. The voltage change can be more clearly seen by connecting an oscilloscope in place of the voltmeter. The display looks something like this:
When the switch is turned to A, the voltage on the supply side of the resistor is 6 V and that on the capacitor side is 0 V. By Ohm’s Law, the current through the resistor is 6/10 0005600 μA. Charging begins and the voltage across the capacitor (see graph) rises steeply. The voltage on the supply end of R1 stays at 6 V, but the voltage at its other end is increasing. The voltage difference across R1 is decreasing. Ohm’s Law still applies, so the current through R1 is decreasing. This means that the rate of charge of C1 is decreasing and the voltage across it rises more slowly.