CrazyEngineers Archive
Old, but evergreen and popular discussions on CrazyEngineers, presented to you in read-only mode.
@Raith • 19 Oct, 2007
I am currently still going to school as an EE major, I am taking an electronics devices class, I seem to be having issues with this diagram:

Sorry if it is sloppily drawn. It asks to Calculate the magnitude and phase angle of the circuit impedence, I know to find the impedence it is 1/(2PI*f*c), which is around 4 ohms. But how does one find the magnitude[convert to V which the formula works as (sqrt. Vc^2 + Vr^2)]. My question is this, would I need to change uF and f into V to get the magnitude or does it work like the magnitude formula but the result would not come out as voltage?
@Kaustubh Katdare • 20 Oct, 2007 Re: Hope this is the right place to ask this

Raith, the most appropriate place for this question is our Electronics & Electrical Engineering section where we discuss everything related to electricity 😀 .

-The Big K-
@Raith • 20 Oct, 2007 You're right but this is for newbies like me haha, I think I figured it out after reading over some sections of the book. Cheers!
@reachrkata • 29 Oct, 2007 Woow !!! need to brush up all the forgotten Basic electricity studied ages back during Engineering.

But If I remember it right :
expressing the impedance as a Complex number
Z = R + jXc where Xc is 1/(2*Pi*f*C)

then i think the magnitude is Sqrt (R^2 + Xc^2)
and the phase i think is tan inverse (Xc/R)

I only vaguely remember this, so people please correct me If i am wrong ???

As i see, there is no need of any voltage. the frequency is given only for the impedance calculation.
@crook • 30 Oct, 2007 You can use subscripts and superscripts using special bb codes. Look https://www.crazyengineers.com/forum/showthread.php?t=1259
@surendarkashyap • 01 Mar, 2008 magnitude is sqrt(R^2+Xc^2)
phase is actuallt arc tan(Z/R)where Z=(Xc-Xl)[Xl is impedence due to inductor which is zero in this case)
so phase(according to this prob )is arc tan(Xc/R)
tats rite!!!!!!
@sauravgoswami • 04 Mar, 2008 Dude,hope u have got your ans,and solve it accordingly..welcome!!!
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