Electronics - Capacitors - Discussion

@Arif-CUTM

As you said;
C=Q/V.
But q =i*t.
And here Q is proportional to C,
Then how can you say current decreases? please explain to me.

The capacitor does not operate based on ohm's law. Capacitors are non-linear devices which means that Voltage and current are not linearly proportional to each other. At t=0 assuming a dc source, i.e. battery is connected to a capacitor, the capacitor will act as a short therefore Vc = 0 and I = maximum. As the charge builds up on the capacitor plates, the voltage difference between the capacitor plates increases (hence Vc increases), so the voltage difference between either terminal of the battery and the plate it is connected to decreases.

So the electric field in the wire decreases. Therefore the current in the wire will decrease in time. Another way to think of it is when the capacitor becomes fully charged, it would essentially become open thus it would not allow current to flow through it effectively limiting the current.

Q=CV
C=Q/V
C=Q/IR by ohm's law V=IR.

When the capacitor is charged then the voltage is decreased, so the current also decreases.

No, because when the cap charge I = suddenly increase and voltage take some time.

When charged, capacitor voltage decreases because the capacitor and voltage is inversely proportionally.

AS we know that Q = CV or C = Q/V.

When the capacitor is charged then the voltage is decreased, so the current also decreaes.

More and more charge are getting into the capacitor as storage.

And so the charge of the source decreases as potential difference lessens. So current flowing into capacitor decreases and also because of a C=Q/V.

@Aswathi.

Here is no resistance. But in ohm's law constant is resistance. So we can't apply ohm's law here. As well as here capacitor is constant.

So we apply C=Q/V. Since here capacitor is charging so when Q increases we decreases. So current also decreases.

C = Q/V.

Capacitance is constant here. When Q increases voltage decreases. So current also decreases.

Because when capacitor is fully charged it becomes equal to the voltage source or battery and when there is no potential difference then current does not flow means I=0.