Impedance of a Capacitor

[flash]

I know the impedance of a capacitor is,

$$\frac {1}{j \omega C}$$

so in an audio circuit it let's more high frequency energy through, which is obvious looking at the equation. When a capacitor is in parallel with a fixed resistor, I worked out the magnitude of the impedance of the pair to be

$$\frac{R \omega C}{\sqrt{R^2 + (\omega C)^2}}$$

and this pair should still have lower impedance at higher frequencies, right?
So my question is, how does this expression reflect this? I just can't see it.

 

[ice109]

I know the impedance of a capacitor is,

$$\frac {1}{j \omega C}$$

so in an audio circuit it let's more high frequency energy through, which is obvious looking at the equation. When a capacitor is in series with a fixed resistor, I worked out the magnitude of the impedance of the pair to be

$$\frac{R \omega C}{\sqrt{R^2 + (\omega C)^2}}$$

and this pair should still have lower impedance at higher frequencies, right?
So my question is, how does this expression reflect this? I just can't see it.

first caps and inductors have reactance, the resistance plus reactance is the impedance of the circuit. just correcting a nomenclature mistake.

second i have no idea how you got that for the total impedance of the circuit

$$V=I|Z|$$

$$|Z|= \sqrt{R^2 + (\chi _c)^2}}$$
$$|Z|= \sqrt{R^2 + (\frac{1}{j \omega C})^2}}$$

as you see with omega in the denominator as freq goes down impedance goes to infinity

 

[flash]

second i have no idea how you got that for the total impedance of the circuit

ahhh i meant parallel, sorry

I got it like this:
$$\frac {1}{Z_eq} = \frac {1}{R} + \frac {1}{j \omega C}$$

$$Z_eq = \frac {R \omega C}{R + j \omega C}$$

$$|Z_eq| = \frac {R \omega C}{\sqrt{R^2 + ( \omega C )^2}}$$

Thanks for your help with this.

edit: should the second term in the first equation be jwC, not 1 on? could be my problem

 

[ice109]

ahhh i meant parallel, sorry

I got it like this:
$$\frac {1}{Z_eq} = \frac {1}{R} + \frac {1}{j \omega C}$$

$$Z_eq = \frac {R \omega C}{R + j \omega C}$$

$$|Z_eq| = \frac {R \omega C}{\sqrt{R^2 + ( \omega C )^2}}$$

Thanks for your help with this.

edit: should the second term in the first equation be jwC, not 1 on? could be my problem

a high pass filter is a cap and a resistor in series though

 

[flash]

The circuit I'm working on has a number of capacitors that can be switched in parallel with a resistor to give different frequency response depending on which one you select.

I reworked the above as:
$$\frac {1}{z_{eq}} = \frac {1}{R} + j \omega C$$

which gives
$$|z_{eq}| = \frac {R}{\sqrt{1 + ( \omega CR)^2}}$$