# Homework Help: AC Circuitsahhh

1. Mar 4, 2005

### airkapp

A 1.20 kΩ resistor and a 6.8 μF capacitor are connected in series to an AC source. Calculate the impedence of the circuit if the source frequency is a) 60 Hz, b) 60,000 Hz

I think I should this equation:

Vrms = Irms, Z : XL = 2ΠfL : XC = 1 / 2ΠfC

Z = √R2 + ( XL – XC )2 so: Irms = Vrms / Z

My main problem here is I don’t know how to do the equation without a given L. Do I need a given L (in Henry’s) to do the problem, and if I don’t, how else can I do it? Any help appreciated,
Thanks,
air

2. Mar 4, 2005

### faust9

First: Read this again "A 1.20 kΩ resistor and a 6.8 μF capacitor are connected in series"

Second: what are the impedance equations?

Third: How do you deal with resistances in series circuits (impedance is analogous to resistance).

Finally: Reread this "Calculate the impedence of the circuit if the source frequency is a) 60 Hz, b) 60,000 Hz". RMS voltage or current are not required for this calculation.

Good luck.

3. Mar 4, 2005

### Curious3141

You only need an inductance if there's an inductor in the circuit. Here there is none.

It's best to deal with impedances as complex quantities since you can handle both magnitude and phase neatly in one step.

The impedance of the capacitor is $$\frac{-j}{\omega C}$$, a purely imaginary value.

The impedance of the resistor is just $$R$$, a real value.

The impedance of the series combination of the two is found just like in a d.c. circuit, by addition, giving a complex value :

$$Z = R - \frac{j}{\omega C}$$

The magnitude of the impedance is found like finding the magnitude of any complex number :

$$|Z| = \sqrt{R^2 + \frac{1}{{\omega}^2 C^2}}$$

The phase is found by getting the argument :

$$\phi = \arctan{\frac{\frac{-1}{\omega C}}{R}} = - \arctan{\frac{1}{\omega RC}}$$

Since $$\omega = 2\pi f$$ where $$f$$ is the given frequency, work out the magnitude and phase in each case, and you're done.

EDIT : Sorry for all the changes in the LaTex, I was having a tough time formatting it properly. It's fixed now.

Last edited: Mar 4, 2005
4. Mar 4, 2005

### airkapp

thanks. That's what I thought about the inductance but I wasn't sure. I'm new to this. So the "i" is just imaginary. Everything else seems to just be math given the equations you gave me.

thanks again.

Last edited: Mar 4, 2005
5. Mar 4, 2005

### Curious3141

Instead of "i" (which can be confused with current), physicists use "j" to signify the square root of negative one.

Complex impedances are like phasors. A purely real impedance means the voltage and current are in phase. A purely imaginary impedance means the voltage and current are 90 degrees out of phase, which leads or lags depends on the sign of the impedance. A complex impedance means voltage and current are out of phase at some other angle.