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Let's say a series circuit has two loads and a battery with 12 volts. Do the loads each use 6V? Or do they each use 12V? And then let's say there is a parallel circuit with a battery (12V) and two loads in parallel. Same question, do they each use 6V or do they each use 12V?enter image description here

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  • \$\begingroup\$ Can you please draw a schematic? It will be clearer \$\endgroup\$ – M.Ferru May 25 '17 at 0:14
  • \$\begingroup\$ Adding a circuit diagram will help so we can understand where you're coming from. \$\endgroup\$ – Envidia May 25 '17 at 0:14
  • \$\begingroup\$ Added diagrams. \$\endgroup\$ – kyrdothien May 25 '17 at 0:19
  • \$\begingroup\$ Do you mean "12 volts" instead of "12 loads"? \$\endgroup\$ – uint128_t May 25 '17 at 0:36
  • \$\begingroup\$ Sorry, yes, I edited it. \$\endgroup\$ – kyrdothien May 25 '17 at 0:51
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you are correct, in the 1st case, both loads are 6V, in 2nd case, both loads are 12V. Such problem can be generally solved by kirchhoff law.

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For your first drawing, with the loads in series, both loads will have the same current. The voltage across each load will depend on that load's resistance. The load with the higher resistance will see a higher voltage. The total of the individual load's voltages will equal the battery voltage.

For your second drawing, with the loads in parallel, both loads get the full battery voltage. The current in each load will depend on its resistance.

The voltage, current, and resistance in any load are related by Ohm's Law: E = I x R (Voltage (electromotive force) = current times resistance).

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  • \$\begingroup\$ For the first drawing, if both loads had the same resistance they would have the same voltage of 6V, correct? And for the second, if both loads have the same resistance, then each would have 12V? \$\endgroup\$ – kyrdothien May 25 '17 at 0:30
  • \$\begingroup\$ In the series case, if both loads have the same resistance, each would see half the supply voltage. In the parallel case, both loads will see the supply voltage, regardless of their resistances. \$\endgroup\$ – Peter Bennett May 25 '17 at 0:59

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