Bob Bruhns bbruhns at erols.com
Sat Aug 20 16:40:47 EDT 2005

```Yes!  That's MUCH better.

Bacon, WA3WDR

----- Original Message -----
From: "John E. Coleman (ARS WA5BXO)"
<wa5bxo2005 at pctechref.com>
Sent: Saturday, August 20, 2005 2:44 PM

> Whoa, Bacon, that ASCII thing doesn't look like a
triangle to me.
>
> But I think this link will so what you wanted to
show.
>
>
http://wa5bxo.shacknet.nu/QuickBridge/Quick%20Bridge%20test.GIF
>
> John
>
>
>
> -----Original Message-----
> [mailto:amradio-bounces at mailman.qth.net] On Behalf Of
Bob Bruhns
> Sent: Saturday, August 20, 2005 11:00 AM
> To: Discussion of AM Radio
> Subject: Re: [AMRadio] AM Audio
>
> Cool!  Here is a crude Ascii-art diagram.  Heh heh,
> yeah, it's supposed to be a triangle.  The base of
the
> triangle is the input voltage, the left side is the
> voltage across the resistor, and the right side is
the
> voltage across the inductor.
>
> This is the situation when the resistance equals the
> inductor impedance.  The angle between the V(R) and
> V(L) is 90 degrees, and you would calculate the
voltage
> compared to V(in) as V(L) = V(R) = V(in) * sin(45
deg)
> = V(in) * 0.7071 because the angle of V(R) and V(L)
> compared to V(in) is 45 degrees.
>
>                             /\
>                       /    90    \
>   V(R)      /                        \      V(L)
>           /                                   \
>      /                                              \
> /__45____________________45__\
>
>                      V(in)
>
>
>   Bacon, WA3WDR (modern artist!)
>
>
> ----- Original Message -----
> From: "Geoff" <w5omr at satx.rr.com>
> To: "Discussion of AM Radio"
> Sent: Saturday, August 20, 2005 11:23 AM
> Subject: Re: [AMRadio] AM Audio
>
>
> > Bob Bruhns wrote:
> >
> > > <>Yes, that will work. But remember, you want
> > > equal voltages across the resistor and the
> inductor.
> > > Surprisingly, this will not be 50% of the applied
> > > voltage, but it will be about 71% of the applied
> > > voltage. This adds up to about 1.414 times the
> > > applied voltage! You should confirm this by
> > > measuring both voltages directly.
> > >
> > > This happens because the impedance of the
> > > inductor is reactive, and the current is equal,
so
> > > the phases ofthe voltages differ by 90 degrees.
> > > This phase shift causes a partial cancellation
when
> > > the voltages are added, so 71 + 71 = 100 instead
> >> of 142. It's interesting to observe.
> >
> >
> > Actually, Bacon, John's entire message was thus...
> > you two agree.  and I present it here, so that
others
> > may learn from this experience, as well.
> >
> > ________________
> >
> > If all you need is a measurement of inductance, the
> easiest thing is to series up a variable resistor
with
> the choke and apply a low voltage from a filament
XFMR
> (6 - 20 volts).  Adjust the resistor so that the
> voltage measured across the choke is equal to the
> voltage across the resistor.  If for instance a 10
volt
> source is used, you will measure about 7 volts across
> the resistor as well as across the choke.
> >
> > You are probably asking why it is not 5 Volts
across
> each and the answer to that is because of the phase
> difference of 45 degrees  when XL = R.  The sine of
45
> degrees is .707 and .707 X 10 Volts is about 7 volts.
> But the main thing is what ever the voltage is, by
> equalizing the voltage across the resistor and choke
> results in the creation of a resistance that is equal
> to XL of the choke at 60 CPS.
> >
> > Measure the resistance of the variable resistor
after
> equalizing the voltage
> > drops and you have the XL of the choke at 60 cps.
> >
> > L = XL / 2 * pi * F
> >
> > L = XL / 6.28 * 60
> >
> > L = XL / 376.8
> >
> > Divide the resistance by 376.8 and you have the
> inductance.
> >
> > ________________________________
> >
> > Thanks for the response.  From all the active
posters
> in here,
> >  I admire and respect the intellect for the
> electronic abilities
> > and talents mostly of  WA3WDR, K4KYV, WA3VJB,
> > KD5OEI and of course, WA5BXO, and not necessarily
> > in that order.
> >
> > --
> > 73 = Best Regards,
> > -Geoff/W5OMR
> >
> >
> >
>
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> > Help: http://mailman.qth.net/mmfaq.html
> > Post: mailto:AMRadio at mailman.qth.net
> >
>
>
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