Electronics - Capacitors - Discussion

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.

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.

@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.

When voltage exists one end of the capacitor is getting drained and the other end is getting filled with charge.This is known as charging. Charging creates a charge imbalance between the two plates and creates a reverse voltage that stops the capacitor from charging. As a result, when capacitors are first connected to voltage, charge flows only to stop as the capacitor becomes charged. When a capacitor is charged, current stops flowing and it becomes an open circuit. It is as if the capacitor gained infinite resistance.

My question is.

Q=CV.

So when charge increases voltage increases.
But voltage is directly proportional to current by ohms law.

Then how the current 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.

C = Q/V.

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

@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.

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.

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.