| Electronics > Projects, Designs, and Technical Stuff |
| Investigations into the stored energy of inductors. |
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| intabits:
I'm confused. --- Quote from: makerman on January 22, 2019, 07:45:23 pm ---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 --- End quote --- Same resistance in both coils, but L2 uses thicker, lower resistance wire. Doesn't that mean that it must have more turns? And if so, doesn't that mean that it must have a higher inductance? Assuming other dimensions are similar. (You don't specify those, and the pic shows more than 2 coils) |
| makerman:
Thanks Intabits, i've posted the inductance values the wrong way around. |
| makerman:
If i connect the large inductor to the DC PSU, the PSU reads 18v , 0.17A so 3.06W, but if i do the calculation for the energy transfer after five time-constants : R = 102 L = 4.65 TC (L/R) = 0.045588235 TCx5 (to reach 99.5% of charge) = 0.22549 Energy = 0.5 x 4.65 x 0.17^2 = 0.067193 Power = (energy/TCx5) = 0.29798439 It indicates only 0.29 watts, what am i doing wrong ? |
| IanB:
--- Quote from: makerman on January 30, 2019, 12:05:18 am ---If i connect the large inductor to the DC PSU, the PSU reads 18v , 0.17A so 3.06W, but if i do the calculation for the energy transfer after five time-constants : R = 102 L = 4.65 TC (L/R) = 0.045588235 TCx5 (to reach 99.5% of charge) = 0.22549 Energy = 0.5 x 4.65 x 0.17^2 = 0.067193 Power = (energy/TCx5) = 0.29798439 It indicates only 0.29 watts, what am i doing wrong ? --- End quote --- Everything? Your power supply is indicating resistive power dissipation. Voltage = 18 V Resistance = 102 Ω Current = 18/102 = 0.176 A Continuous dissipated power = 18 x 0.176 = 3.17 W So far this has nothing to do with inductance. It is only about resistance. Then you try to do a calculation with inductance. You suggest the time constant is L/R = 4.65 H / 102 Ω = 0.046 s ? But I don't think that calculation works. Does henrys divided by ohms give seconds? (Yes, it does. However, you are looking at a transient accumulation of energy in a magnetic field compared to a continuous dissipation of heat in a resistor. You are not comparing like with like.) |
| makerman:
Thanks Ian, i have a lot to learn! |
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