Re: [Phys-l] Ugh - RLC circuit issues

[Brian Whatcott <betwys1@sbcglobal.net>]

At 10:41 AM 5/19/2007, Josh, you wrote:

>   I'm on my last nerve with a couple of student independent lab project
> that won't behave - perhaps the combined wisdom of phys-l can guide me to
> the light at the end of the tunnel!
>
>   #1- they have a crystal radio kit, and they're trying to identify the
> frequency of the radio station that they're picking up.  I suggested that
> they'd get a good answer by determining the inductance of the coil and the
> capacitance of the variable cap at the correct tuning.  For this, they're
> trying to find the L by pairing it with a known C and finding the resonant
> freq., and vice-versa for the C.  The problem is in the measurement; I'm
> getting some inconsistent data.  Just by number of coils, area, and
> length, the L should be in the 300 uH range, but I haven't been able to
> get an experimental value for the inductance anywhere near that.  I'm
> driving the unknown coil, a known electrolytic cap (100-1000 uF range)
> with a ScienceWorkshop freq. generator, and continuously measuring the
> voltage across a resistor in series with the cap and coil, looking for the
> highest voltage difference to find the highest current.  Any ideas?

In principle, there's nothing wrong with this style of measurement -
your choices of test components may not be helping however.

If the variable tuning cap were  say 400 pF on down,
and the radio frequency were between 400 kHz and 4 MHz,
what size of inductance would one in fact expect?

f = 1/ [2.pi sqrt( L.C) ]
so 4E5 = 1 / [2.pi. sqrt ( L. 4E-10)]
and L = 300 uH (about)

The quality factor (Q) of this tuned circuit is intended to pass
the audio sidebands of the transmitted signal - say 400kHz/20 kHz
or Q=20, but the Q of your test version if I understand it,
being proportional to root L/C   is rather low because the  C is
now about one million times bigger. This would constitute an invisibly
small  tuned resonance amplification factor.   Q is factored with 1/R
so resistance in the circuit further drops the Q (if that were possible!)

>   #2- another circuits project, a la Mythbusters this time.  They're
> looking for the "break even" time for a fluorescent light bulb
> to make up its high energy start-up cost in low steady-state consumption.
> We trying to measure the current drawn with a PasPort current probe.  The
> max current for these is listed at 1.1 A, and the steady state current for
> the 20W tube should be .167 A, but the start-up current is apparently high
> enough to spark, fry, and let the smoke out of the probe...  I have analog
> meters and a scope as well, but nothing that can capture the I vs t curve
> to integrate. Ideas?
>
> Thanks for the help!
> Joshua Gates
> Stoneleigh-Burnham School

Here is a better application for a small dropper resistor.
The safe arrangement might be a switched power line to an isolating
 1:1 transformer and a few ohms in one of its secondary lines to
check the transient voltage change on fluorescent startup.
 Playing with line voltage has its risks of course.
 One might make an issue of using a wooden lab bench, rubber mat,
rubber gloves etc.

 Some of these fellows may be called to mend power line breaks
 on a wet stormy night, in future years.



Brian Whatcott    Altus OK    Eureka!