In theory, adding a wire (with no resistance) between the positive and negative terminals would produce an infinite current. In this way, a device can draw what ever current it needs.

However, in a series circuit with a resistor placed before the device, do we have a maximum current that the device can draw (which may be lower than what the device needs, even though the voltage is correct)?

Example, suppose a single resistor in a circuit reduces the current to 5Amps, would a device attached after the resistor be limited to 5Amps?


closed as unclear what you're asking by pipe, PeterJ, Voltage Spike, Daniel Grillo, winny Dec 20 '17 at 15:48

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\$ Well, you can't break Ohm's law and conservation of energy. That said a way of cheating your question would be using a very efficient buck converter to get higher currents (at lower voltage). \$\endgroup\$ – Wesley Lee Dec 15 '17 at 3:05
  • \$\begingroup\$ What limitations do you want to put on the "device". Could it be, for example, a fully charged 100V battery? \$\endgroup\$ – Spehro Pefhany Dec 15 '17 at 3:11
  • 1
    \$\begingroup\$ Could the device be a (synthetic) -4.99 ohm resistor? \$\endgroup\$ – Spehro Pefhany Dec 15 '17 at 3:55
  • \$\begingroup\$ three things: 1) there is no after the resistor ... the current flow in a simple series circuit is the same at every point in the circuit .... 2) the resistor reduces the maximum current ... you cannot get higher current than what flows when the resistor is the only device .... 3) the wire is an "attached device", what is the current in the wire? \$\endgroup\$ – jsotola Dec 15 '17 at 4:42
  • \$\begingroup\$ @SpehroPefhany, I'm assuming a normal device like an MCU. \$\endgroup\$ – Edward Dec 15 '17 at 13:24

If you connect a 1 Ohm resistor across a 5 volt power supply, 5 Amps will flow through the resistor. If you then connect some other device, say, another 1 Ohm resistor, in series with the original resistor, the current through both resistors will now be 2.5 Amps. You now have 5 volts across 2 Ohms, so Ohm's Law says the current will be 2.5 Amps.

You can't say "the resistor reduces the current to 5 Amps", and assume that 5 Amps will be available to whatever you might connect in series with the original resistor, because the current in a circuit depends on the total resistance in the circuit.

  • \$\begingroup\$ Sure, but I guess my point is the current will never exceed 5 Amps (which is effectively its limit). Could there be a case where a device specifies it needs x volts (after the resistor), but the current can never be as high as is needed (because of the preceding resistor)? \$\endgroup\$ – Edward Dec 15 '17 at 13:05
  • \$\begingroup\$ An unusual example: a load specifies it's required voltage as 6 Volts and 100 Amps. A resistor is placed in the circuit to reduce the 12 Volts power supply to 6 Volts. At this point, could the load ever draw as much current as it needs to operate? \$\endgroup\$ – Edward Dec 15 '17 at 14:48
  • \$\begingroup\$ The voltage dropped by a resistor depends on the current through it. For your 6V/100A case, you would need a 0.06 Ohm resistor to drop 6 volts. If the load draws actually draws only 50 Amp for a moment, the resistor will only drop 3 volts, so the load will get 9 volts. If the load instead attempts to draw 200 Amp, the resistor will attempt to drop 12 volts, so there would be no voltage left for the load. In practice, most things will vary their current demand depending on the applied voltage, so it is hard to say what the actual load voltage and current will be in either of these cases \$\endgroup\$ – Peter Bennett Dec 15 '17 at 16:25
  • \$\begingroup\$ I presume a load has no concept of voltage - it is only concerned with the current passing through it? \$\endgroup\$ – Edward Dec 15 '17 at 16:29
  • \$\begingroup\$ @Edward: Voltage and current are intimately related - see Ohm's Law. \$\endgroup\$ – Peter Bennett Dec 15 '17 at 17:35


Refer back to ohms law, Voltage = Current * Resistance.


There's nothing like "preciding" or "after". A resistor placed "preciding" or "after" a load ( or device as you say), will limit the current flowing through it. Total current flowing is NOT dependent ONLY on that resistor. It will depend also on the added load's resistance, which is in series. Anyway that current will always be less than the current, when no load was present, but only the resistor. Hence its called current limiting resistor.


Not the answer you're looking for? Browse other questions tagged or ask your own question.