Yes! That's MUCH better. Bacon, WA3WDR
----- Original Message ----- From: "John E. Coleman (ARS WA5BXO)" <[EMAIL PROTECTED]> To: "'Discussion of AM Radio'" <[email protected]> Sent: Saturday, August 20, 2005 2:44 PM Subject: RE: [AMRadio] AM Audio > 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----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] 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" <[EMAIL PROTECTED]> > To: "Discussion of AM Radio" <[email protected]> > 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 > > > > > > > _______________________________________________________ > _______ > > AMRadio mailing list > > Home: http://mailman.qth.net/mailman/listinfo/amradio > > Help: http://mailman.qth.net/mmfaq.html > > Post: mailto:[email protected] > > > > _______________________________________________________ _______ > AMRadio mailing list > Home: http://mailman.qth.net/mailman/listinfo/amradio > Help: http://mailman.qth.net/mmfaq.html > Post: mailto:[email protected] > > > > _______________________________________________________ _______ > AMRadio mailing list > Home: http://mailman.qth.net/mailman/listinfo/amradio > Help: http://mailman.qth.net/mmfaq.html > Post: mailto:[email protected] >
