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I hear people say things like "I only put 5 amps through the circuit but I put a bunch of volts". I don't understand how this is possible if V=IR. Lets say you have a circuit with 5 ohms of resistance so V=I(5). The amount of voltage and current I am allowed to put through it has to be proportional.

Can anyone can give a good intuitive answer (don't go too in depth with math) that is understandable?


marked as duplicate by Ignacio Vazquez-Abrams, PeterJ, Daniel Grillo, Scott Seidman, Kaz Jan 15 '16 at 16:42

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  • \$\begingroup\$ Transformers don't "just" exchange volts and amps. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 15 '16 at 5:32
  • \$\begingroup\$ Everything obeys Maxwell's equations, but these require calculus and vector math and are pretty complicated. So we use a simpler model called lumped constant where we assume resistors, voltage sources, etc. connected by ideal wires. For DC or steady-state circuits we also assume the voltage and currents are constant. This is where we use Ohm's law and KVL and KCL. But transformers don't work at DC, they require alternating current. AC analysis is more complicated; energy is still conserved but peak voltage and peak current may happen at different times. \$\endgroup\$ – MarkU Jan 15 '16 at 5:37
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    \$\begingroup\$ There are all kinds of circuits that are not resistors. Ohm's Law only applies to resistors and resistances. \$\endgroup\$ – Daniel Jan 15 '16 at 5:50
  • \$\begingroup\$ So are voltage and amperage always proportional in a DC circuit with resistance? \$\endgroup\$ – Rhezner Jan 15 '16 at 5:53
  • \$\begingroup\$ Voltage and current are always in the ratio of the resistance. If the resistance changes, then the ratio changes. There are devices with variable resistance - any wire for instance has a slight temperature coefficient of resistance - though thermistors have a much much greater variation. \$\endgroup\$ – Neil_UK Jan 15 '16 at 6:23

If you re-read your question, then it sort of answer's itself. Yes...in a pure DC circuit, Ohms Law is King, and so V=I*R. But you stated they said "I only put 5 Amps through the Circuit....so if they put a constant 5 Amps through the circuit, then the voltage will be the current times the resistance. Since they didn't state the Resistance, then the value of the Voltage may be a "bunch of Volts"....if the resistance is high enough. Simply put....under normal everyday simple DC circuits....most people drive them with either a Voltage or Current Source...but typically not both. If they drive the circuit with a Current, then they will see the Voltage go up proportionally. So I believe maybe they didn't mean they "Put In" both Current and Voltage, but they put in Current and Measured a Bunch of Voltage...or Vice Versa.

  • \$\begingroup\$ I think I understand, I probably sound like an idiot but I love to understand things intuitively. Thanks for answering guys! \$\endgroup\$ – Rhezner Jan 15 '16 at 17:56

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