PHY- capacitance

Why not write up the processes in mathematics.
Let Vs be the supplied voltage to charge up the capacitor
Vc be the capacitor voltage
I be the charging current
R be any resistance in the circuit, including that of the connecting wires
Hence, by Ohm;s law, we have
Vs = Vc + IR
i.e. I = (Vs-Vc)/R
But Vc = Q/C, where Q is the charge on the capacitor and C is its capacitance.
Thus, I = (Vs- Q/C)/R
During charging, Q gradually increases, the term (Vs-Q/C) will decrease (because Vs and C is constant), thus I will decrease with time.
The current I will be reduced to zero when Q increases to a level such that Q/C = Vs, the capacitor is then fully charged.
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The same equation also applies on discharging just by putting Vs = 0 volt
we have, I = - Q/RC
[ the - ve sign only means that the current through R now flows in opposite direction to that during charging].
Very clearly, when Q is gradually decreasing, I will decrease at the same time until all charges on the capacitor has gone.

Consider the capacitor is being charged by a constant voltage supply.

When it is being charged, there are charges accumulating on the metal plates of the capacitor. By Q = CV, there will be a voltage increase across the capacitor.

Positive charges are accumulated on the plate connecting to the positive terminal of the voltage supply, and the negative charges to the negative terminal.

So, there will be a potential difference increase in the capacitor, with the direction opposite to the direction of e.m.f. of the voltage supply.

Hence, the voltage in the circuit decreases.

As time goes by, the number of charges increase, resulting in the gradual decrease in the voltage in the circuit, hence, the current decreases gradually.


When the capacitor is discharging, it is isolated from any voltage supply.

It is connected to an electrical load. Hence, positive charges will move away from the positive metal plate of the capacitor.

As time goes by, the voltage across the capacitor drops and hence the current decreases also.


When the capacitor is being charged, the current in the circuit, I = I0e-t/RC, where I0 is the initial current, R is the resistance in the circuit, C is the capacitance. So, it decreases with time.

When it is discharging, the current in the circuit, I = I0e-t/RC, where I0 is the initial current, R is the resistance in the circuit, C is the capacitance. So, it decreases with time.



2009-04-18 09:55:09 補充：
Yes, the voltage across the capacitor increases. But the directions of potential difference across the capacitor and the voltage supply are opposite. So, voltage in the circuit decreases. Therefore, current decreases.