COP 20.00 (2000%) Times, Reactive Power Energy Source Generator,

[Farmhand]

Also just wondering, since both half cycles are identical can we not just disregard one half cycle and assess the circuit in
a DC situation using only the positive half cycle ? To simplify the power analysis.

..

P.S. I looked at the power consumed by the capacitors and things seem a bit odd to me then, anyone explain the values at the bottom of this sim ? It says 178 Watts consumed by the resistor and it also shows 78 Watts consumed by each capacitor. 

Code.

$ 1 5.0E-6 15.472767971186109 50 5.0 50
v 384 208 384 144 0 1 60.0 177.0 0.0 0.0 0.5
w 384 144 432 144 0
w 432 208 384 208 0
c 656 112 656 176 0 9.999999999999999E-6 -7.954571889469782
c 720 176 720 240 0 9.999999999999999E-6 -7.954571889469783
159 656 176 720 176 0 0.1 1.0E10
159 720 176 720 112 0 0.1 1.0E10
159 656 240 656 176 0 0.1 1.0E10
w 656 240 720 240 0
w 656 112 720 112 0
w 768 352 768 144 0
w 688 192 688 288 0
T 432 144 512 208 0 7.0362 1.0 -1.2230422064702071 1.2921222355024904 0.999
r 544 112 624 112 0 100.0
w 624 112 656 112 0
w 544 112 512 112 0
w 512 112 512 144 0
w 512 208 512 240 0
w 512 240 656 240 0
w 736 144 768 144 0
w 672 352 768 352 0
w 672 208 672 288 0
w 688 288 688 416 0
150 576 416 672 416 0 2 0.0
150 576 352 672 352 0 2 5.0
w 672 416 688 416 0
w 576 368 576 432 0
w 464 336 576 336 0
R 448 336 384 336 1 2 120.0 2.5 2.5 -1.0471975511965976 0.33333300000000005
R 448 400 384 400 1 2 120.0 2.5 2.5 -3.490658503988659 0.33333300000000005
x 283 345 362 351 0 24 Charge
x 255 406 363 412 0 24 Discharge
152 496 432 576 432 0 2 5.0
w 576 400 480 400 0
w 480 400 480 448 0
w 496 448 480 448 0
w 448 400 480 400 0
w 464 336 464 416 0
w 464 416 496 416 0
w 448 336 464 336 0
w 672 288 672 352 0
o 0 32 0 289 160.0 1.6 0 -1
o 0 32 1 291 320.0 9.765625E-5 1 -1
o 13 64 1 35 320.0 9.765625E-5 2 -1
o 3 64 1 35 80.0 9.765625E-5 3 -1
o 4 64 1 35 80.0 9.765625E-5 4 -1


..

[Farmhand]

If this is in fact legitimate, then I want to be using it.  My positive thinking tells me that if it will work identically on each half cycle of AC it should work with DC as well so I tried the sim with a diode by itself in the supply side and also with a capacitor there as well as
the diode.

If true and we can apply the principal with DC and batteries and capacitor banks then the batteries should show a net charging current.

[SchubertReijiMaigo]

Also just wondering, since both half cycles are identical can we not just disregard one half cycle and assess the circuit in
a DC situation using only the positive half cycle ? To simplify the power analysis.

..

P.S. I looked at the power consumed by the capacitors and things seem a bit odd to me then, anyone explain the values at the bottom of this sim ? It says 178 Watts consumed by the resistor and it also shows 78 Watts consumed by each capacitor. 

The sim display only peak watt not the integrated value per unit of time...
The white curve is an instantaneous I*U curve, but unfortunately no integrating function for the curve.
You can look visually the area of the white (power) curve above or under the zero line and compare with input.
Or doing yourself integration if you know math...

[Jdo300]

Hi Stefan,

Here's the modified simulation using the resistor series/parallel switching. I haven't spent any time tuning the component values, but here's the basic template for everyone to play with.

$ 1 5.0E-6 3.5993318835628396 40 5.0 50
v 320 240 320 176 0 1 60.0 177.0 0.0 0.0 0.5
w 320 176 368 176 0
w 368 240 320 240 0
c 640 144 640 208 0 9.999999999999999E-6 -173.65305112063913
c 704 208 704 272 0 9.999999999999999E-6 -173.65305112063913
159 640 208 704 208 0 0.1 1.0E10
159 704 208 704 144 0 0.1 1.0E10
159 640 272 640 208 0 0.1 1.0E10
w 640 272 704 272 0
w 640 144 704 144 0
w 752 320 752 176 0
w 672 224 672 320 0
T 368 176 448 240 0 4.0 1.0 0.15665227448245278 0.19767468597969223 0.999
r 480 80 544 80 0 50.0
w 608 144 640 144 0
w 480 144 448 144 0
w 448 144 448 176 0
w 448 240 448 272 0
w 448 272 640 272 0
w 720 176 752 176 0
w 656 320 752 320 0
w 656 240 656 320 0
w 672 320 672 448 0
150 560 448 656 448 0 2 0.0
150 560 384 656 384 0 2 5.0
w 656 448 672 448 0
w 560 400 560 464 0
w 448 368 560 368 0
R 432 368 368 368 1 2 120.0 2.5 2.5 -1.0471975511965976 0.33333300000000005
R 432 432 368 432 1 2 120.0 2.5 2.5 -3.490658503988659 0.33333300000000005
x 267 377 346 383 0 24 Charge
x 239 438 347 444 0 24 Discharge
152 480 464 560 464 0 2 5.0
w 560 432 464 432 0
w 464 432 464 480 0
w 480 480 464 480 0
w 432 432 464 432 0
w 448 368 448 448 0
w 448 448 480 448 0
w 432 368 448 368 0
w 656 320 656 384 0
r 544 144 608 144 0 50.0
w 608 144 608 80 0
w 480 144 480 80 0
159 480 144 544 144 0 0.01 1.0E10
159 544 144 544 80 0 0.01 1.0E10
159 544 80 608 80 0 0.01 1.0E10
w 560 112 784 112 0
w 784 112 784 448 0
w 784 448 672 448 0
w 576 96 752 96 0
w 752 96 752 176 0
w 512 160 512 320 0
w 512 320 656 320 0
o 0 32 0 35 261.87124863169134 5.237424972633828 0 -1
o 0 32 1 291 523.7424972633827 9.765625000000001E-155 1 -1
o 13 32 0 35 149.65776766268445 2.993155353253689 2 -1
o 13 32 1 35 279.96809277222553 9.765625E-105 3 -1

NOTE: The two graphs on the right side show the voltage and power dissipation through only one of the two resistors. Notice how the voltage on the return spike appears to be taller, but it is only because of the series/parallel switching. If you look at the current through the entire resistor network, it looks similar to the first simulation.

- Jason O

[vasik041]

Model and simulation results in LTSpice