FW: Three Voltages and Series Resistor - Revisited - (was - Magnetic quenching.)

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Original poster: "John H. Couture" <couturejh-at-mgte-dot-com> 


Corrected posting ---

I have finally completed the algebra for this problem. The correct numbers
for the two triangles are shown below. The numbers that I showed in the
previous post were ballpark numbers because I did not have the time to do
the algebra. Note that there has been a lot of posts about wattmeters
lately. Wattmeters to find the input watts of a TC are not required anymore.
The tests below will find the input watts. However, I have not had time to
try the Three Voltages and Series Resistor Test so I hope some coilers will
give these tests a try.

If you want to avoid the math I am working on a graph to give you watts
directly without math. With controlled sparks you will now be able to find
the energy (joules) per spark using the information below. However, the
exact number of sparks per second will be required. Maybe Marco will give us
this method.

John Couture

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-----Original Message-----
From: John H. Couture [mailto:couturejh-at-mgte-dot-com]
Sent: Tuesday, March 16, 2004 11:08 PM
To: Tesla list
Subject: RE: Three Voltages and Series Resistor (was - Magnetic
quenching.)



Ed, Jim, Robert, All -

This problem caught my attention so I dug further. I at first thought that
it would not be possible to find the unknown load reactance without using a
capacitor or inductor.
I found that only one resistor is needed and three voltage readings and no
capacitor or inductor.
This will also give you input watts without a wattmeter for your Tesla coil.

Here is a problem to work on -
Given 3 voltages (60 Hz) and a series resistor
Find the resistance and inductance of the load.

Volts across load & res   = 120
Volts across load (TC?)   = 108
Volts across resistor     = 15
Series resistor ohms      = 15

There is more than one way to solve this problem. In fact there is one
method that is very easy. Just find the sides of two right angle triangles.
One is the triangle with the load and series resistor and the other the
triangle with only the load.

I have simplified the problem somewhat by using a 15 ohm resistor and 15
volts across the resistor for a one amp current in the circuit. Try the
problem with a 10 ohm resistor and 20 volts across the resistor.

The load and resistor triangle -
    x= 98.7    y = 68.25    z = 120

The load triangle -
    x = 83.7   y = 68.25    z = 108

The load resistance is 83.7 ohms
The inductance is 68.25/377 = .181 = 181 mh

Can you fill in the calc details? Find the power factor? Input Watts?
Can you solve the problem in another way?
I wonder how Ed's students would solve this problem now?
I wonder if I got it right?

John Couture

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