WHEN YOU CONNECT IT THERE WILL BE big inrush into the cap (sorry about typing here)......and then there will be no current at all. As no dc can flow in a capacitor.
But you can see this with maths if you want...do V/Xc...., AND you will see that you get zero current when freq=o (dc).
But then you see that this is not quite right, because there will be small leakage current...ah well...
A capacitor charges according to I = C dv/dt if it is being charged with a constant current source. You are not doing this. You are putting a voltage source with (effectively) infinite current output across the capacitor. the current will be (theoretically) infinite as the capacitor immediately charges to the voltage output by the voltage source. Once the capacitor has charged, the charge current will be zero.
What stops this in real life is the Effective Series Resistance (ESR) of the capacitor that limits the inrush current. Also, the voltage source will have an output resistance that also limits the current
If you replace the capacitance C with a resistor R, then it's a simple application of Ohm's Law. V=IR, or since you are looking for current with a known voltage and resistance, use the equivalent form I=V/R
Ok 1 case is we are using C 47uF
With voltage like rectifer of transformer so current will depend on load.
The c charge in few mili sec.
After charging the capcitor is removed from rectifier and load is connted led with few ohm R resistance.
Then current and voltage will fall slowly.
How to get graph or value it is droping of both V and I
Like discharging of C with R load