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In a system involving a power source that charges a battery, and a load that subsequently runs from the battery, are we limited to 25% efficiency at best?

Since charging a battery (or capacitor) loses 50% of the power transferred to heat, and by the same logic, another 50% while discharging into a load...does that not imply that battery systems of this type cannot be more than 25% efficient?

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    \$\begingroup\$ Where are you getting these figures (50% of power "transferred" to heat)? \$\endgroup\$ – Shamtam Nov 16 '17 at 20:39
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    \$\begingroup\$ YOu are going to have to quote your source for those numbers... \$\endgroup\$ – Trevor_G Nov 16 '17 at 20:39
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    \$\begingroup\$ I believe it is about the recently discussed capacitor paradox. electronics.stackexchange.com/questions/337487/… \$\endgroup\$ – Eugene Sh. Nov 16 '17 at 20:42
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    \$\begingroup\$ Anyway, look here at the "solvability" section. \$\endgroup\$ – Eugene Sh. Nov 16 '17 at 20:51
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    \$\begingroup\$ Me thinks your numbers only apply if you completely discharge and recharge each time. Which for a battery results in buying a new one... \$\endgroup\$ – Trevor_G Nov 16 '17 at 20:56
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Not at all, for batteries that approximate capacitor behaviour (Which is to say nearly none of them) you charge with a current source not a voltage source, which is potentially very efficient.

Batteries that behave as a voltage source in series with a resistor (Where the resistor value appears to vary) you charge with a current limited constant voltage.

Now on discharge, maximum power transfer occurs when RSource = RLoad, but we don't want maximum power! We want (somewhat close to) maximum energy, so we make sure the batteries internal resistance is much smaller then the apparent resistance of the load.

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  • \$\begingroup\$ I knew I was missing a key concept here, and the way you answered has filled it in for me. \$\endgroup\$ – evildemonic Nov 16 '17 at 21:06

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