Electronics > Projects, Designs, and Technical Stuff
Investigations into the stored energy of inductors.
makerman:
Hello, i'll be detailing my main project in this thread.
I'm investigating the stored energy in an inductor, the standard equation says it's : W=1/2LI^2, implying that the two ways of increasing the stored energy are by increasing either the inductance or the current, i'm looking into what happens when we have two inductors, both with the same resistance but one with significantly higher inductance, will the latter store more energy given the same power input?
I've done my first coil-pair test :
Coil 1.
Wire diameter : 0.53mm
Resistance per metre : 0.0775 Ohms
Inductor resistance : 24.5 Ohms
Self-inductance : 0.77 Henries
Coil 2.
Wire diameter : 0.71mm
Resistance per metre : 0.0432 Ohms
Inductor resistance : 24.5 Ohms
Self-inductance : 0.232 Henries
I'm using a pulse width modulator to pulse the coil, and a pair of diodes on the coil output so that only the CEMF spike powers the load, the load being a Neon bulb for visual comparison of the coils' output. This isn't ideal since it's visual judgement so i'm formulating a better way to compare output now that i have a decent scope.
In the test (pulsing at 300Hz with a 50% duty cycle) the first coil lit the Neon and the power usage of the PWM was 1.45 Watts.
The first coil was then swapped out for the second, which lit the Neon visibly brighter and used 0.93 Watts.
I'm not sure why the higher-inductance coil uses less power, i'm thinking it may be due to the larger time-constant ?
The next step is to do a comparison of two coils with a much larger difference in inductance, i've got the large-inductance coil (over 3.5kM of 0.85 wire!) which has an inductance of 4.5 Henries and a resistance of 102 Ohms and i've ordered a kilo of 0.4mm wire which i'll cut to a resistance of 100 Ohms then perform the same experiment.
Picture of the large inductor attahced, it's a beauty :)
duak:
The time constant of an R-L circuit is L/R. (This is in contrast to an R-C circuit where the time constant is R*C.)
Are you familiar with exponential functions and how they apply to R-L and R-C circuits?
If you go through all the math, given the same applied voltage and the same series resistance, the larger inductor will charge to a lower current for the same length of time. Since power is V * I, a lower current implies lower power.
Cheers,
makerman:
Thanks duak, exponential functions yes, how they apply to circuits no but i shall read up on it.
I'm puzzled as to why the bulb shone more brightly with less power in?
IanB:
--- Quote from: makerman on January 22, 2019, 10:09:45 pm ---I'm puzzled as to why the bulb shone more brightly with less power in?
--- End quote ---
(Power Out) = (Power In) − (Power Losses)
How might you account for and measure the power losses in each experiment?
makerman:
--- Quote from: IanB on January 22, 2019, 10:22:35 pm ---(Power Out) = (Power In) − (Power Losses)
How might you account for and measure the power losses in each experiment?
--- End quote ---
Thanks IanB.
Well these are air-cored so no core losses, therefore only losses in the windings which is dependent on the resistance and both coils have the same resistance, hence my confusion as to it drawing less current but seemingly delivering more power to the bulb. Neons need a realtively high startup voltage but a smaller maintaining voltage and the current is on the order of a few mA or less.
As for measurement, i just got my first decent scope a few days ago and will learn how to measure power in/out this week.
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