Calculating Impedance of 2.80k W Resistor and 4.60mF Capacitor

[purduegirl]

Homework Statement

A 2.80-k W resistor and a 4.60-mF capacitor are connected in series across a 46.0-Hz AC generator. What is the impedance of the circuit?

Homework Equations

Z = sqrt ( R^2 + (XL -XC)^2)

Xc = 1/wC = 1/ 2*pi*46.0 Hz* 4.60uF
XL = wL

The Attempt at a Solution

I can seem to figure out how to find inductance L without knowing the current. Any ideas?

 

[Defennder]

Is it a 2.8kW resistor or 2.8k ohms resistor? And there isn't any inductor in the circuit, is there? It looks like you have to solve some DE in order to get the current. Just setup the voltage drop across the capacitor and resistor to be equivalent to the source. But it appears a little tedious. I think there's a graphical way of doing this by drawing phasor diagrams,but I'm a little rusty with drawing phasors. You can also solve this using complex numbers if you've learned complex impendances.

 

[dlgoff]

With no inductors in the circuit, XL=0. Just plug in the values you have to find Z. Note your units for capacatance needs to be in farads to find XC.

 

[purduegirl]

So I would take the $$\sqrt{R^2 + Xc^2}$$

Xc = 1/2*Pi*(4.60E-6 F)
Xc = 2.89E-5 F
Xc^2 = 8.35E-10 F


So using the math, I would get $$\sqrt{7840000 Ohms + 8.35E-10 F}$$
$$\sqrt{.0065464}$$
0.0809 Ohms*F

 

[purduegirl]

This isn't the correct answer. Where am I going wrong?

 

[alphysicist]

So I would take the $$\sqrt{R^2 + Xc^2}$$

Xc = 1/2*Pi*(4.60E-6 F)

In the original post, the problem statement said the capacitor was 4.6 mF, not 4.6 $\mu$F. Was the original post incorrect?

Also, you are missing the frequency f in this expression.

Xc = 2.89E-5 F[/itex]

These numbers are not the result of the above calculation. In the formula for $X_C$, the C is in the denominator; but it looks to me like when you put it in your calculator the C went in the numerator.