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Re: Meter Shunts



Original poster: BunnyKiller <bunikllr@xxxxxxx>

Hey Tom...

what application do you have in mind for the meter?? do you want to measure current? what voltages are you applying to the meter?

most meters use about 50mV on average to make a full swing ( from 0 to the other side) if you are using the meter for current, then a shunt will be needed. If you are measuring voltage, a single resistor on one leg of the line is needed... but the resistor needs to be sized for the maximum voltage you intend to measure and this resistor works in tandem with the resistance of the meter ( 2 resistors in series gives a voltage drop between the 2 resistors thus the low voltage to feed the meter)

Scot D



Tesla list wrote:

Original poster: Thomas DeGregorio <tommacs@xxxxxxxxxxxxx>

Thank you very much for the equation

Anyway yhat I find rather odd is that I called up a store and ask if I can get a meter and then shunt it. The meter will be for measuring the current, but the before the shunt I would use two resistors to drop the current evenly since the meter can't handle the whole load. This is odd because he said this cannot be done, why is that? The person didn't explain to me why either, I tihnk they didn't know.
Anyway before I was told that I don't need to use a shunt to measure the voltage nor resistors, but wouldn't the current have some effect on the meters coil?
Thanks,
Tom


On Feb 16, 2005, at 10:03 AM, Tesla list wrote:

Original poster: "Steve Conner" <steve.conner@xxxxxxxxxxx>

>Does anyone know the exact equation to find the new current from a
>capacitor? This equation in particular I can not seem to find anywhere

I'm not surprised. I'm an EE and I've never heard the phrase "new current"
used in this kind of context before.


I can only assume that you got it from this page

<http://users.tm.net/lapointe/HowItWorks.htm>http://users.tm.net/lapointe/HowItWorks.htm


which was one of the first hits that Google produced for "capacitor new current"


In this case, the answer is related to the "surge impedance" of the tank circuit. If you charge a capacitor C to a voltage V, then connect it to an inductor L, the peak value of the oscillating current I will be:

I=V/Z where Z=sqrt(L/C)


This applies to a single lossless LC resonator. If you want an exact answer for a Tesla coil with losses and two coupled circuits, you have to get back to basics and solve the differential equations describing the system. We normally use a simulator like Orcad PSpice for this rather than doing it by hand.

But in practice, the surge impedance formula gives a reasonable answer for
the peak primary current in the first half cycle, so it is good enough for
estimating peak current stresses on your capacitors and such like.


Steve C.