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Sci.Electronics.Basics -> Help using complex numbers to work out LRC circuits ???
There are 7 messages in this thread.
You are currently looking at messages 1 to 7.
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Author: jalbers@bsu.eduDate: 15:48 19-06-08
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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 mH inductor, and 100 pF capacitor
wired in series are connected to a 1V AC 10Mhz power supply.
Z =3D 50 + 62.8j =96 159j
Z =3D 50 =96 96.2j or 108.417@-62.53
Suppose that I wanted to find the voltage drops across L, R, and C
when the power source is at .707V or 1@45 or .707 + .707j .
I =3D V/Z or
I =3D 1@45 / 108.417@-62.53 =3D .0092@107.53 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) * (.0092@107.53) =3D .46@107.53 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) * (.0092@107.53) =3D .577@197.53 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) =3D 1.4628@17.53 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
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Author: Greg NeillDate: 16:28 19-06-08
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<jalbers@bsu.edu> wrote in message
news:e52da699-2f16-49f1-9b8a-f877cf0cf3c0@w1g2000prd.googlegroups.com=20
> 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...
>=20
> Suppose that a 50 ohm resistor, 1 mH inductor, and 100 pF capacitor
> wired in series are connected to a 1V AC 10Mhz power supply.
>=20
> Z =3D 50 + 62.8j =16 159j
> Z =3D 50 =16 96.2j or 108.417@-62.53
Your 62.8j Ohms doesn't look right for a 1 mH inductor
at 10 MHz. I think you "lost" the power of ten multiplier.
Should be more like j62.8K .
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Author: jalbers@bsu.eduDate: 20:09 19-06-08
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On Jun 19, 4:28=A0pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
> <jalb...@bsu.edu> wrote in message
>
> news:e52da699-2f16-49f1-9b8a-f877cf0cf3c0@w1g2000prd.googlegroups.com
>
> > I am trying to understand XLR circuits and the use of complex numbers
> > to find solutions. =A0I 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 mH inductor, and 100 pF capacitor
> > wired in series are connected to a 1V AC 10Mhz power supply.
>
> > Z =3D 50 + 62.8j =A0 159j
> > Z =3D 50 =A0 96.2j or 108....@-62.53
>
> Your 62.8j Ohms doesn't look right for a 1 mH inductor
> at 10 MHz. =A0I think you "lost" the power of ten multiplier.
> Should be more like j62.8K .
The 1 mH means 1 micro henry, not 1 milli henry.
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Author: John FieldsDate: 20:35 19-06-08
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On Thu, 19 Jun 2008 17:09:46 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:
>On Jun 19, 4:28 pm, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
>> <jalb...@bsu.edu> wrote in message
>>
>> news:e52da699-2f16-49f1-9b8a-f877cf0cf3c0@w1g2000prd.googlegroups.com
>>
>> > 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 mH inductor, and 100 pF capacitor
>> > wired in series are connected to a 1V AC 10Mhz power supply.
>>
>> > Z = 50 + 62.8j 159j
>> > Z = 50 96.2j or 108....@-62.53
>>
>> Your 62.8j Ohms doesn't look right for a 1 mH inductor
>> at 10 MHz. I think you "lost" the power of ten multiplier.
>> Should be more like j62.8K .
>
>The 1 mH means 1 micro henry, not 1 milli henry.
---
No, it doesn't. In standard scientific notation, the lower case 'm'
stands for 1/1000, not 1/1000000.
In order to eliminate confusion you should either use the symbol 'µ'
or spell out 'microhenry' if that's what you mean.
JF
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Author: jalbers@bsu.eduDate: 20:04 20-06-08
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On Jun 19, 8:35=A0pm, John Fields <jfie...@austininstruments.com> wrote:
> On Thu, 19 Jun 2008 17:09:46 -0700 (PDT), "jalb...@bsu.edu"
>
>
>
>
>
> <jalb...@bsu.edu> wrote:
> >On Jun 19, 4:28=A0pm, "Greg Neill" <gneill...@OVEsympatico.ca>
wrote:
> >> <jalb...@bsu.edu> wrote in message
>
> >>news:e52da699-2f16-49f1-9b8a-f877cf0cf3c0@w1g2000prd.googlegroups.com
>
> >> > I am trying to understand XLR circuits and the use of complex number=
s
> >> > to find solutions. =A0I think that I am applying and interpreting th=
e
> >> > 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 mH inductor, and 100 pF capacitor
> >> > wired in series are connected to a 1V AC 10Mhz power supply.
>
> >> > Z =3D 50 + 62.8j =A0 159j
> >> > Z =3D 50 =A0 96.2j or 108....@-62.53
>
> >> Your 62.8j Ohms doesn't look right for a 1 mH inductor
> >> at 10 MHz. =A0I think you "lost" the power of ten multiplier.
> >> Should be more like j62.8K .
>
> >The 1 mH means 1 micro henry, not 1 milli henry.
>
> ---
> No, it doesn't. =A0In standard scientific notation, the lower case 'm'
> stands for 1/1000, not 1/1000000.
>
> In order to eliminate confusion you should either use the symbol '=B5'
> or spell out 'microhenry' if that's what you mean.
>
> JF- Hide quoted text -
>
> - Show quoted text -
I didn't know how to generate the "mu" character and even if I figured
it out I was afraid it wouldn't show up correctly on everyone's
computer so I used m for micro forgetting that m stands for milli like
ma in milli amps. I will spell it out in the furure to avoid any
confusion.
What about the rest of the math that follows ?????????????
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Author: Charlie SiegristDate: 20:09 20-06-08
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On Thu, 19 Jun 2008 19:35:14 -0500, John Fields wrote:
> On Thu, 19 Jun 2008 17:09:46 -0700 (PDT), "jalbers@bsu.edu"
> <jalbers@bsu.edu> wrote:
>>> Your 62.8j Ohms doesn't look right for a 1 mH inductor at 10 MHz. Â I
>>> think you "lost" the power of ten multiplier. Should be more like
>>> j62.8K .
>>
>>The 1 mH means 1 micro henry, not 1 milli henry.
>
> ---
> No, it doesn't. In standard scientific notation, the lower case 'm'
> stands for 1/1000, not 1/1000000.
>
> In order to eliminate confusion you should either use the symbol 'µ' or
> spell out 'microhenry' if that's what you mean.
To OP: for ease of use in usenet posting, 'u' will do for 'micro', e.g.
'1uH'. The message will be understood. FWIW, when I saw mH, I *assumed*
microhenry, just because it makes more sense, both through understanding
of the construction of inductors, and the application at 10MHz. Still,
it's never a good thing to make assumptions in electric circuits, and
using 'm' for 'micro' is certainly incorrect.
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Author: Greg NeillDate: 00:26 21-06-08
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