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Re: capacitance bridge problem



At 08:28 PM 10/22/00 -0700, John Mallinckrodt wrote:

In the arrangement shown in the figure, a potential difference V
is applied, and C1 is adjusted so that the voltmeter between
points b and d reads zero. This "balance" occurs when C1 = 4.00
microF. If C3 = 9.00 microF and C4 = 12.0 microF, calculate the
value of C2.

______________________
| | |
| V ===C1 ___ ===C4
----- b|____/ V \___|d
--- | \___/ |
| ===C2 ===C3
|_________|____________|

The "obvious" answer you can do in your head:

______________________
| | |
| V ===4 ___ ===12
----- b|____/ V \___|d
--- | \___/ |
| ===3 ===9
|_________|____________|


This is obvious by symmetry. You don't even need to know the formula for
the impedance of a capacitor; even if you make the all-too-common mistake
of thinking the impedance is the reciprocal of what it really is, you
stumble onto the right answer anyway.

I learned something tonight, something that I previously "knew"
but apparently did not "understand." The lesson was delivered
after scratching my head for way too long over the [...]
seemingly straightforward problem (which I had made the mistake of
assigning before solving) from the current edition of a well known
intro text. I thought others might enjoy the challenge as well.
Here's the problem (what's the lesson?)

I don't know what the lesson is, or what the challenge is, except to wonder
why anybody would put such a foolish circuit into a textbook.

a) The circuit doesn't work if the voltmeter is too good; there is no
provision for removing "historical" charge from the nodes b and d.

b) The circuit doesn't work if the voltmeter is too crummy; the voltmeter
will discharge nodes b and d before you have a chance to take the
reading. And that's not the only things that could go wrong; this circuit
seems tailor-made to amplify voltmeter nonidealities that would go
unnoticed in other applications.

c) Anybody with an IQ greater than room temperature (centigrade) would
replace the battery with an AC signal generator, and use an AC voltmeter.

Here's why this is not nitpicking: When I teach circuit analysis, I make a
big fuss about what I call "the principle of reasonability". When
analyzing a circuit, the student should not treat it as a pile of
components thrown together helter-skelter; the thing was designed for a
purpose, and if you know the purpose it is a great guide to the analysis.

When a homework assignment violates the principle of reasonability, that's
a bad thing.

=======================

We now move from the elementary circuits course to the real-world physics lab:

d) Anybody who was serious about building a capacitance bridge would use a
lock-in amplifier, so that the detector was synchronous with the excitation.

e) Anybody who really cared would _not_ use an adjustable capacitor, but
would use a ratio transformer instead:

______________________
| | |
| | $
| === Std $ constant
| | $ inductance
_|_ | ___ |
/exc\ b|____/det\___|d
\___/ | \___/ |
| | |
| | $ adjustable
| === Unknown $ inductance
| | $ on same core
|_________|____________|

That's because the transformer ratio depends on turns ratio, which can be
constructed in terms of known quantities (integers) much more easily than
capacitance ratios or resistance ratios can be.

With such a setup, you can take a 1pF capacitor and measure it to 1 part
per million (+- 1 femtoFarad). That's in a 1Hz bandwidth; you can do even
better if you care to do a little signal averaging, and if you throw the
Std capacitor into a jug of liquid nitrogen so it doesn't drift.

Not only are such circuits not covered in introductory courses, you could
easily get an undergrad degree in physics or electrical engineering without
seeing such a thing, or even seeing any mention of lockins or ratio
transformers. But in a real research lab, such things are as common as
mice in meadow.