rajasekhar
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Posts: 7
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Your calculations are correct. The smaller the voltage charged to the cap, the lesser the energy lost in the resistor. You can consider each step charging voltage is V/N, replete this N time. When N-->inf, you would see no energy loss. Another simple way is to check the current waveform. In your example-charging directly to V, would result in impulse current of VC. If you do the two steps charging you would see CV/2 impulses two times. I hope it is helpful.
Thanks, Raj.
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