LRC omplex Numbers (Try #2)

Check your numbers for the inductor.

For some reason all of my complex numbers are bing treated as email addresses. The numbers informt of the @ are being turned into ... . I have replaced the @ symbols with an & symbol so that this will not happen again. Everwhere yo see an & it should be an @ symbol.

Can you give me more of a hint as to what is wrong with the inductor. Is the error in the instantaneous voltage across the inductor, the value of the inductor, or the XL calculation?

-----

Z =3D 50 + 62.8j =96 159j Z =3D 50 =96 96.2j or 108.417&-62.53 deg

Suppose that I want to find the voltage drops across L, R, and C when the power source is at .707V =3D 1&45 deg =3D .707 + .707j .

I =3D V/Z or I =3D 1&45 deg / 108.417&-62.53 deg =3D .0092&107.53 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: (50&0 deg) * (.0092&107.53 deg) =3D .46&107.53 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: (62.8&90 deg) * (.0092&107.53 deg) =3D .577&197.53 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: (159&-90) * (.0092&107.53 deg) =3D 1.4628&17.53 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.

You should not be looking at just the real part of the voltage across each part. You're using _complex_ currents, and _complex_ impedances; why should the voltages not be complex also?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

news: snipped-for-privacy@t54g2000hsg.googlegroups.com

The voltage angle looks wrong. I make it out to be about=20

-160 degreees (I'll let you calculate an accurate value).

Sounds like the source is 1 volt peak, not 0.707.

The RLC doesn't see "the real part" of the voltage, they see the real voltage, namely 1 volt peak. The load has no idea that you consider the applied voltage to be at 45 degrees, since it has no way of knowing your time reference.

The same AC voltage can be named 1 at 0 degrees, 0.707+j0.707, or

0+j1.00; it just depends on where you're standing. A voltmeter will read exactly the same for all cases, so the RLC can't tell the difference. If you define the voltage source as having a non-zero angle, you can *interpret* the resulting complex load current as different angles, but the final magnitudes must be the same.

John

1V*(cos(q) + j*sin(q)) =3D (0.707 + j*0.707)V for q =3D 45 deg

It seems that he's looking for the instantaneous voltage drops across the passive components when the driving=20 voltage is 45 degrees into its cycle.

Yes, that is "exactly" what I am trying to do. Can this be done the way I am attempting to go about it?

xt -

+197.53 deg is the same as 162.47 deg which is close to your -160 deg.

Am I correct in regards to just looking at the real part of the voltages if I am interested in an instantaneous voltages across X, L, and R, since my calculations are based on the instance when Vsource =3D .

707V =3D 1&45 deg =3D .707 + .707j . Not that doing so has any practical value. I just can't seem to find a "real" scope or DVOM that displays imaginary numbers :)

news: snipped-for-privacy@z66g2000hsc.googlegroups.com

news: snipped-for-privacy@t54g2000hsg.googlegroups.com

Okay.

=3D

Sure.

Yes, but you can't throw away the imaginary part of the voltage, just because it's imaginary!

John

I

to

rs

he

en

It sure is funny how the sum of the instataneous voltage drops (just the "real" parts) any place in the AC cycle in my calculations always adds up to the source which is a good thing because Kirchoff's voltage law holds up.

Let V the voltage source be 1&x deg where 0

Not funny at all, but exactly as expected. The sum of all the imaginary voltages in the RLC also add up to the imaginary part of the source. That is because the sum of all the voltage around any loop has to be zero, regardless of whether that voltage is expressed as a sum of components at right angles (real and imaginary) or by any other description.

Now sum the imaginary components, then do the complex math to sum the whole voltages around the loop. All versions have to add up to zero.

--
Regards,

John Popelish