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I have a series circuit with a 12VDC source with a 60Ah capacity. I can calculate power loss by having the initial wattage that is drawn from the battery ( 120W) subtracting by the final wattage(star @ 4.8 W) -> 115.2W Power loss. My question is whether that remaining 4.8 W of power goes back to the battery to charge the battery or if that power is lost or what happens to the remaining power given that if the resistance of the wire is negligible and if the resistance of the wire has some value. Also, if the star was replaced by plasma (resistance much lower than wire), would the resistance decrease from 30 ohms to a lower value? Circuit with battery and resistors.

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    \$\begingroup\$ A 12 V battery driving a 30 ohm load does not deliver 10 A, initially or at any other time. From Ohm's Law, it produces only 0.4 A initially. What did you read or hear that made you think it would produce 10 A? Please add this information to the question so we can address why it's wrong. \$\endgroup\$
    – The Photon
    Dec 9, 2021 at 3:32
  • \$\begingroup\$ Wait, so is the current from each resistor all going at 0.4A? \$\endgroup\$
    – JuicyYellow
    Dec 9, 2021 at 3:35
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    \$\begingroup\$ Yes, they are in series, so the current must be the same in all of them. \$\endgroup\$
    – The Photon
    Dec 9, 2021 at 3:36
  • \$\begingroup\$ current from each resistor ... resistors do not generate current ... it should say current through each resistor \$\endgroup\$
    – jsotola
    Dec 9, 2021 at 5:58

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A) The power drawn from the battery isn't 120W, it's 4.8W.

B) There is no "remaining power".

C) Plasma doesn't have "resistance much lower than wire".

D) No, you have 30 ohms of resistors, the resistance will not be lower than 30 ohms.

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