Electrical Engineering - Capacitors - Discussion

Initially, as given in the problem the capacitor is in uncharged condition.at the instant when the switch is closed the voltage across the capacitor is zero because the capacitor only charges when an infinite amount of current flows through it. So, at the time of starting not much amount of current flows through ckt and hence, the voltage across the capacitor is 0
We know;

C = Q/V.
V = Q/C.

And also Q=idt at the starting there is no change in current with respect to time due to that there is no change accumulation.

Voltage across the capacitor cannot change immediately since, it requires infinite current to flow. This is physically impossible. (i=C dv/dt). Since, before closing the switch voltage across the capacitor is ZERO and at the instant switch has been closed also voltage across the capacitor is ZERO.

C does not allow the sudden change in voltage so here at t = (0-) C have no charge it is given in question then at t = (0+) C behave like a short circuit and we know that short circuit have 0 voltage.

The capacitor does not allow sudden changes in voltage, therefore, the voltage across the capacitor is zero. Similarly, Inductor does not allow sudden changes in current.

Actually voltage across capacitor can not change instantaneously so in this change case answer is zero.

Q=CV,
V=Q/C,
Because capacitor is uncherged soQ=0 c.
So V=Q/C= 0/C =0V.

Here it's using DC source (battery) so the circuit will act like open.

The capacitor rating is independent of voltage hence it's zero.

Yes, the capacitor takes some time to charge.

For DC source, capacitor act like open.