# Charging and discharging capacitors - current time graph

by jsmith613
Tags: capacitors, charging, current, discharging, graph, time
Mentor
P: 10,759
 Quote by jsmith613 well no but it does change the circuit diagram
Equivalent circuits are a powerful analysis tool. If you're worried about esthetics, post the original circuit diagram and we can work from that
P: 614
 Quote by gneill Equivalent circuits are a powerful analysis tool. If you're worried about esthetics, post the original circuit diagram and we can work from that
see diagram!!
Attached Thumbnails

Mentor
P: 10,759
 Quote by jsmith613 see diagram!!

The circuit in the diagram has a small problem: There's no resistance in the path when the switch is in the 'charging' position. For ideal components that would mean that the battery would "see" no resistance between its own p.d. and the current p.d. of the capacitor, which begins at zero volts. The ideal battery would supply an infinite current to bring the capacitor p.d. up to that of the battery instantaneously. No gradual increase or decrease in current in that case, just BANG! You're done!

In real life there is always resistance in the path, even if it's just the resistance of the wiring. A small modification to your circuit will fix this. Two choices: Either move the resistor so that it will be in either current path, or add a second resistor to represent the resistance in the charging path:

For both versions, when the switch is moved to the left to charge the capacitor, the resulting circuit is the same as the Charging Circuit that I posted earlier. Similarly, the circuit that results when the switch is moved to the right is the same as the Discharging Circuit in that post.

In any circuit, current is moved by a potential difference. The greater the potential difference the greater the current for a given resistance in between. This is just Ohm's Law: I = ΔV/R.

What you need to do is spot the initial potential difference that each circuit begins with (which devices in the circuits are holding a potential before the path is closed?). That sets the initial current. Then determine how that potential difference changes over time as the current flows.
Attached Thumbnails

P: 614
 Quote by gneill ---
ok so is this what your saying:

Initially there is a greater potential difference between the resistor, power source and capacitor.

When charging the p.d decreases so the flow of charge (the current decreses)

when the circuit discharges the potential difference between the resistor and capacitor is maximum at the start. As charge leaves the capacitor the p.d falls and therefore so does the current?
Mentor
P: 10,759
 Quote by jsmith613 ok so is this what your saying: Initially there is a greater potential difference between the resistor, power source and capacitor. When charging the p.d decreases so the flow of charge (the current decreses) when the circuit discharges the potential difference between the resistor and capacitor is maximum at the start. As charge leaves the capacitor the p.d falls and therefore so does the current?
Yup. That's it
P: 614
 Quote by gneill Yup. That's it
yey
thanks for this

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