Impedance angle in RL and RC circuits

[StuartSpencer]

Homework Statement

This isn't so much of a problem itself but I found it interesting that on the All About Circuits website, two of the questions give a phase angle for impedance as opposite the angle of current. I understand the current is 90 degrees out of phase with voltage in both RL and RC circuits, but do not understand the reasoning behind the impedance phase angles.

Here are the relevant links,

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-3/series-resistor-inductor-circuits/

Homework Equations

V = IR [/B]

The Attempt at a Solution

Not sure how to attempt a solution here, I looked online and used various searches but didn't find anything. If anyone could shed some light on this it would be appreciated.

 

[gneill]

Why don't you present one of the examples here for discussion so that helpers don't have to read through both pages to find what you're referring to?

 

[StuartSpencer]

Why don't you present one of the examples here for discussion so that helpers don't have to read through both pages to find what you're referring to?

Alright, in the RC circuit link there is an RC series circuit with

R = 5 Ohms
C = 100 microFarads
V = 10V / 60Hz

The total impedance is calculated to be 27 ohms and an angle of -79 degrees.
The total current is calculated to be 370.5 mA at an angle of +79 degrees.

In the RL circuit,

R = 5 ohms
L = 10 mH
V = 10V/ 60Hz

The total impedance is calculated to be 6.262 ohms at an angle of +37 degrees
The total current is calculated to be 1.597 amps at an angle of -37 degrees.

In both examples the current is the opposite angle of the impedance. Like I said I understand how the current differs in phase angle from the voltage, but do not understand where the relationship between phase angle of current and phase angle of impedance comes into play.

 

[gneill]

The current is given by Ohm's law:

$I = \frac{V}{Z}$

Since the voltage source is supplying the phase reference angle, it's angle is 0°. In the equation the impedance is in the denominator, so if its angle is, say Φ, when the division is performed the resulting phase angle of the current is given by 0° - Φ = -Φ. That's basic complex math.

 

[StuartSpencer]

Thank you very much, that clears it up for me!