The Amateur Extra class question of the week comes from sub-element 5 (electrical principles) section C (polar coordinates) [E5C06] In polar coordinates, what is the impedance of a network consisting…

Amateur Extra: Network Impedance

Amateur Extra: Network Impedance

The Amateur Extra class question of the week comes from sub-element 5 (electrical principles) section C (polar coordinates) [E5C06]

In polar coordinates, what is the impedance of a network consisting of a 100-ohm-reactance capacitor in series with a 100-ohm resistor?

A. 121 ohms at an angle of -25 degrees
B. 191 ohms at an angle of -85 degrees
C. 161 ohms at an angle of -65 degrees
D. 141 ohms at an angle of -45 degrees

We need to remember how to plot polar coordinates when it comes to impedance.  First we must plot the given points on an X-Y graph.

Polar coordinates

http://commons.wikimedia.org/wiki/File:Coordonnees_polaires_plan.png

We plot pure resistance on the horizontal, or X axis.  Then we plot the reactive component on the vertical or Y axis.  But we have to remember that capacitive reactance needs to be plotted on the negative.  The result is a point at coordinates (100,-100) on the graph, on the above image, it would be at point M, (only point M would be BELOW the X axis, not above!)  (I need to get a proper graph of this, if you read this message, please remind me!)

to find the magnitude of the impedance, we employ Mr. Pythagoras’ thereom as so:

r = \sqrt{(x^2 + y^2)}

or

r = \sqrt{(100^2 + (-100)^2)} = \sqrt{(10000 + 10000)} = \sqrt{20000} = 141.42

we could bother with using trigonometry to find the angle, but since both components are of the same magnitude, we know that its 45°.   Actually, since it’s below the X axis, its actually negative, or -45°.

Therefore, the answer is D. 141 ohms at an angle of -45 degrees

 

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