Accurate Measurements on pulsed system's harder than you think.

[verpies]

In your sentences hereunder, can you please elaborate some more why we have  to multiply first and average after ?

Because mathematically:  AVERAGE(i₁, i₂) * AVERAGE(v₁, v₂) <> AVERAGE(i₁*v₁, i₂*v₂).
Take a calculator or a spreadsheet and see for yourself.

[EMJunkie]

Why do you not see the problem in your own calculations Chris?

@Tinman - My calculations are in agreeance with what you are seeing in the circuit. Are they not?

I think you will find, if you read TK's post, he is saying the same thing (More power more Light, less power less light), his math is likely more precise than mine:

OK, are we all done chasing the Red Herrings now?

Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)

With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms.  And the power being dissipated by the bulb is I²R = (0.082)² x 34.1 = 0.224 Watt.

With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V.  So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I²R = (0.163)² x 38.8 =  1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.

Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.

Please check my math and assumptions, and insert your own values for input current, V_drop across the bulb, and duty cycle of the pulse train.

Less power is applied in total across components 1 and 2 when the Cap is connected compared to when the cap is disconnected. The Current goes up but the voltage goes way down. Less power Less light on the globe.

There is nothing in this circuit that is out of the Norm. Look at the Tesla Switch... What's the Potential Voltage Difference between the Load?

   Chris Sykes
       hyiq.org

[verpies]

OK, by Ohm's Law and going through my calculations above, if you multiply CH1 x CH2 , and CH1 is the voltage drop across the bulb, and CH2 is the current going to the bulb, you do get the Power dissipated in the bulb.

...but not the power delivered to the rest of the circuit through that bulb.

[woopy]

Because mathematically:  AVERAGE(i₁, i₂) * AVERAGE(v₁, v₂) <> AVERAGE(i₁*v₁, i₂*v₂).
Take a calculator or a spreadsheet and see for yourself.

Youp problem here, excuse my ignorance

If on the same pulsed circuit, my scope shows  for example  :  on Chanel 1, an average (mean)  voltage of 10 volts and on channel 2, an average (mean) current of 2 A, can i simply assume that the average (mean) power is the multiplication of these 2 average (mean) values, and in this case,  the  MEAN POWER of the pulse  is  10  x 2 = 20 watts ? yes or not ?

Laurent

[EMJunkie]

Just popping this one in here.

Magnets increasing the efficiency of a pulsed system.
Can you argue with this one?.

https://www.youtube.com/watch?v=tVNABy8fSlI


Brad

@Brad and All

EXCELLENT Demonstration!!!

EVERYONE should watch this and pay very close attention!

Faradays Law of Induction clearly Assists the action in this System!!!

   Chris Sykes
       hyiq.org