Q:

Need to understand AC current resistor/capacitor circuit

I'm very confused about capacitor & resistor in AC current circuits, I followed the equations but still confused, my question is will the voltage & current be decreased ?

If we applied an AC current 220v/60Hz on point A let's assume R = 1K  C = 100nF, What will voltage and current be on point B ?
Please if you don't mind to let me understand the result by running equations

A:

simulate this circuit – Schematic created using CircuitLab
There is a lot to clarify here: 

This is a series circuit so the current will be the same everywhere
You can't have a current at a point, current flows through something.  Think of water in a pipe current is the rate the water moves through it. This is not a perfect analogy but should give you the idea
Similarly you can't have voltage at a point you have a voltage or potential difference between two points. I know you will be able to point to text books that say the voltage at point 'C' is but that's only a convenience when the reference point can be assumed. Most people here would assume the reference was point 'B' in this circuit as I have marked it with a 'GND' symbol but the correct thing to say would be "the voltage at point 'C' with respect to point 'B' is".

Now if we want to know the current in this circuit we calculate
$$I = \frac{V}{Z}$$ 
When calculating \$ Z \$ we must take into account that current in a capacitor 
lags the voltage by \$90^o\$ so it is not just a case of adding them up.
\$ R = 1k \$,      \$ X_c = \frac{-j}{2 \cdot \pi \cdot 60 \cdot 100n} \approx -j \cdot 26.525k \$
We can use Pythagoras to find \$ Z = \sqrt{R^2 + X_c^2} \approx 26.54k\$
So the impedance is dominated by the capacitor and the current will be lagging the voltage by almost 90 degrees.
I'll leave the rest as an exercise.

