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# Electric circuits - phasor analysis problem watch

1. Question: An a.c supply of amplitude 5 mA and frequency 1 kHz is connected across a parallel combination of a 2 kΩ resistor and a 100 nF capacitor (an RC parallel circuit). Draw a phasor diagram for the circuit and find

a. the impedance of the circuit

I can't seem to get the right answer for this bit (supposed to be 1.25 kΩ). My phasor diagram has V as the reference phasor (horizontal), the capacitor current leading V by 90 degrees, the resistor current in phase with V and the applied current, I = 5 mA at an angle of Φ to it (I'd attach an image, but having problems using The GIMP).

If I use Pythagoras;

I^2 = (I_R)^2 + (I_C)^2 (I_R and I_C being the currents of the resistor and capacitor)

=> I^2 = (V/R)^2 + (VwC)^2 (w = omega, for some reason, unicode char map won't let me copy an omega)

=> I^2 = V^2(1/R^2 + w^2.C^2)

=> V^2/I^2 = (1/R^2 + w^2.C^2)^-1

Impedance, Z = V/I = (1/R^2 + w^2.C^2)^-1/2

If I plug in the numbers, I don't get 1.25 kΩ.. have I done something wrong?

Thanks.
2. Hmm ... shouldn't the impedance be complex?
3. (Original post by shiny)
Hmm ... shouldn't the impedance be complex?
Yeah, there is complex impedance, but I've not got up to that yet.. just revising phasor analysis and there's no complex impedance in it (we did that later on).
4. I get what you get.
5. I worked out what I did wrong.. I used frequency instead of angular frequency, oops.
6. (Original post by Nylex)
I worked out what I did wrong.. I used frequency instead of angular frequency, oops.
I was using 50Hz instead of 1kHz! Oops! yes, it is 1.25kohms
7. (Original post by shiny)
I was using 50Hz instead of 1kHz! Oops! yes, it is 1.25kohms
Lol . Thanks again. Right, now have to go and revise complex impedance. Fun.

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