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

## Help with LRC, Complex Numbers ???

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

check your answers using either a circuit analysis program or

another calculation technique. The circuit analysis program is

the easiest to do.

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

check your answers using either a circuit analysis program or

another calculation technique. The circuit analysis program is

the easiest to do.

LEDs vs Bulbs, LEDs are winning.

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