Rms current in circuit with capacitor resistor and rms output

[a_ferret]

Homework Statement

A 39.5 uF capacitor is connected to a 47.0 ohm resistor and a generator whose rms output is 25.2 V at 60.0 Hz. Calculate the rms current in the circuit.
(also asks for voltage drop across resistor, capacitor and the phase angle for the circuit, but I mostly want to get the first one first)

Homework Equations

I=current, V=voltage, R=resistance, f=frequency

Irms=1/sqrt(2)*Imax
Irms=delta Vc, rms/Xc
Xc=1/(2pi*f*c)
delta Vc,rms=Irms*Xc

and then I start going in circles

Other formulas that might be appropriate:
Vmax=Imax*Z
Z=sqrt(R^2+(Xl-Xc)^2)
Xl=2pi*f*L
L=?? not given in this problem
Power average=Irms^2*R

The Attempt at a Solution

combining lots of these formulas trying to come up with the correct Vrms has proven unsuccessful. I came up with 17.819 A, .37515 A, .21739 A, none of which were right. None of them were successful and I don't remember or care to explain how I got them precisely because I wasn't confident in those anyway.
Overall I still haven't been able to link my given information directly to the answer I need

 

[rock.freak667]

Formulate the expression for Z, then you know that I = V/Z.

 

[a_ferret]

but what do I use for L while setting it up for Z? I don't have it directly in the given information and I don't think I can derive it. Should I just use a default value like 1 or 0?

ps thanks for a response :D

 

[collinsmark]

Calculate the impedance Z of the RC network.

Z = R + \frac{1}{j \omega C}

(true for this particular problem)

Z = R +j \left( \frac{-1}{\omega C} \right)

For this problem, the resistance is R, and the reactance is -1/(ωC).

In general,

Z = \Re \{ Z \} + j \Im \{ Z \}

Use the Pythagorean Theorem to find the magnitude of Z.

|Z| = \sqrt{\left( \Re \{ Z \} \right)^2 + \left( \Im \{ Z \} \right)^2}

And finally [the complex version of] Ohms law to find the current.

i_{RMS} = \frac{v_{RMS}}{|Z|}

 

[a_ferret]

What is j for those equations?

 

[collinsmark]

What is j for those equations?

Here I used the symbol j to represent \sqrt{-1}. This is common in electrical engineering courses, since the symbol i is already taken, representing current.