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One would be inclined to think the average current on resistor R would be exactly the same as in the load L, but I'm not so sure of that fact, due to the effect of the (big, 1000uF+) capacitors. During a significant portion of the AC cycle, the current supplying the load comes from the second capacitor, not the resistor.

enter image description here

What are the calculations involved to assess R's wattage? (Oh, and apologies for the pathetic drawing.)

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    \$\begingroup\$ I am 47 years old and I want to assemble a PSU for an old PSU-less Apple //c I bought cheap off eBay. Also, I don't have any children of appropriate age for that question. Thank you for your concern. \$\endgroup\$
    – JCCyC
    Oct 3 '12 at 21:32
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    \$\begingroup\$ The schematic is actually making my eyes hurt. You've created the Comic Sans of schematic capture. \$\endgroup\$
    – The Photon
    Oct 3 '12 at 22:22
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    \$\begingroup\$ It took me a good half minute to figure out that the thing in the middle was a bridge rectifier and not some crazy Christmas ornament. \$\endgroup\$
    – Nate
    Nov 3 '12 at 6:01
  • \$\begingroup\$ The voltage over R will change with I, which means that the voltage over L will change with the power usage of L. This seems like a bad characteristic for a power supply. Did you consider using a regulator instead of the resistor? \$\endgroup\$
    – Willem
    Oct 4 '16 at 14:03
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The current through R will be the sum of the current through the second filter cap and the load.

To calculate R's wattage, it would be the voltage across R times this current.

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  • \$\begingroup\$ ...squared, of course. \$\endgroup\$
    – JCCyC
    Oct 3 '12 at 22:17
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    \$\begingroup\$ No, I did W = I * V above so there's no square. You can also do I^2*R if that's what you are referring to. \$\endgroup\$
    – Oli Glaser
    Oct 3 '12 at 22:23
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After powering on, and the big cap has charged, there will be no "extra" power dissipated in the resistor. If the second cap supplies current during part of the AC cycle, it will be recharged during the other part of the cycle, through R, so the net effect is the same as if you didn't consider the cap in the first place.

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