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I struggle to understand how if I increase the resistance in a circuit, it requires less current, based of ohms law.

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closed as unclear what you're asking by winny, RoyC, Finbarr, Sparky256, Renan Mar 5 '18 at 16:41

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Ohms law does not speak of temperatures only volts, amps and ohms. \$\endgroup\$ – Andy aka Mar 1 '18 at 9:24
  • \$\begingroup\$ you mean how does this formula work? I=V/R with Resistance formula is: R = ρL/A Where: ρ: Resistivity constant of the material, in Ω.m L: Length of the wire, in meter A: Cross sectional area of the wire, in m^2 R: Resistance, in ohms (Ω) \$\endgroup\$ – Sunnyskyguy EE75 Mar 1 '18 at 9:33
  • \$\begingroup\$ Argh! Resistance blocks the flow of current. Current that does flow through resistance creates heat, even if a tiny bit. \$\endgroup\$ – Sparky256 Mar 4 '18 at 5:18
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I = V/R This means that if you increase R, the circuit will have less current, but it doesn't mean the circuit will require less current. If you have any component that requires some kinda of current to work and u keep increasing your R, the component will not work the same way. So you have to calculate your R, based on how much current the circuit will need to work. Hope it clarifies you.

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  • \$\begingroup\$ Good answer I guess for an ambiguous question. I would suggest you look what the question was before the edit and try to answer that. PS. Who is this "u" person you're talking about? \$\endgroup\$ – Oskar Skog Mar 2 '18 at 0:51
  • \$\begingroup\$ missed that, was referring to the person who asked but i changed it for "the circuit" \$\endgroup\$ – Luís Filipe Mar 2 '18 at 11:11
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Picture a river flowing down a hill. The amount of water that flows is your current. The depth of the water is our voltage. Now build a damn across half of it. What happens?

The water meets more resistance at the damn so less water is flowing out.

(What then happens is that water backs up forming a lake. The depth of the water rises till the current is the same going in as going out. Rivers are a constant current source. Same thing in electronics. If you want the current to remain the same when the resistance increases you have to increase the voltage too.)

Maybe the misunderstanding is that you "need" less current. Think of it this way: Current is what you get for putting in Volt. You get less current.

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