## Help with LRC, Complex Numbers ???

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jalbers
Posts: 18
Joined: Tue Dec 18, 2001 1:01 am
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### Help with LRC, Complex Numbers ???

I am trying to understand XLR circuits and the use of complex numbers to find solutions. I think that I am applying and interpreting the
math correctly but would feel more comfortable if someone with more experience took a look at what I am doing...

Suppose that a 50 ohm resistor, 1 uH inductor, and 100 pF capacitor wired in series are connected to a 1V AC 10Mhz power supply.

Z = 50 + 62.8j â€“ 159j
Z = 50 â€“ 96.2j or [email protected] deg

Suppose that I want to find the voltage drops across L, R, and C when the power source is at .707V = [email protected] deg = .707 + .707j .

I = V/Z or
I = [email protected] deg / [email protected] deg = [email protected] deg or -.0027+.0087j
Taking the real part of -.0027+.0087j means that when the voltage source is at .707 volts, the current through L, R, and C is at -.0027 amps.

The voltage across R:
([email protected] deg) * ([email protected] deg) = [email protected] deg or -.138 + .438j
Taking the real part of -.138 + .438j means that when the voltage source is at .707 volts, the voltage across R is at -.138 volts.

The voltage across L:
([email protected] deg) * ([email protected] deg) = [email protected] deg or -.550 - .173j
Taking the real part of -.550 - .173j means that when the voltage source is at .707 volts, the voltage across L is at -.550 volts.

The voltage across C:
([email protected]) * ([email protected] deg) = [email protected] deg or 1.394 + .440j
Taking the real part of 1.394 + .440j means that when the voltage source is at .707 volts, the voltage across L is at 1.394 volts.

-----

Any comments or corrections would be greatly appreciated. Thanks

MrAl
Posts: 3862
Joined: Fri Jan 11, 2002 1:01 am
Location: NewJersey
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Using complex numbers is a very good way to calculate quantities
in AC circuits so i have to encourage you to continue to pursue
this as you will find it very rewarding.

You got pretty far in your example, and except for one detail
you did very well if you ask me.

The only problem is that when you calculate the quantity across
a circuit element you have to complete the calculation by
taking the amplitude of the complex quantity and perhaps calculate
the phase angle too. You have done this very well with the
other quantities so i see no reason why this should present any
new problem for you. The amplitude as you know is the square
root of the sum of the squares of the imag part and real part,
not simply the real part. This is true for a voltage across
a circuit element also.

For example:
The voltage drop across the resistor R is:
vR=R*V/Z
which comes out to a complex quantity as you already calculated,
then after calculating the amplitude you get:
VR=0.461
with a phase angle of
107 degrees.

The only thing different i did than what you did is i calculated
the amplitude of the voltage across the resistor after multiplying
R*V/Z (a complex quantity) and this is of course equal to:

amplitude(vR)=sqrt(realpart(vR)^2+imagpart(vR)^2)

and this is the peak voltage you would measure with a meter
across the resistor if you had a meter that could read up that
high in frequency, or the peak scope waveform if you have a scope.

As a rule, the amplitude is always calculated LAST, after all
the complex math is done for a given circuit element or node
or branch current.

I have a feeling you will apply this as soon as you read this
post as there isnt much new to you here.

As far as the phase angle goes, since there is only one source
the phase angle can be taken as zero, calculate all the phase
angles, then add 45 degrees to all the answers, or you can
use the complex quantity for the voltage source as you already
noted would be correct.

Lastly, if you are working on something important you should