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I'm using this app called iCircuit for Android (I'm asking here because the iCurcuit forum has been forsaken) and when I have a circuit with a 3V battery and an LED, it lists the current as 2.15GA. I'm new to electronics so I'm wondering why its giving me this because I have connected an LED to a 3V battery before and its been fine. enter image description here

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    \$\begingroup\$ .. if you don't have a current limiting resistor, then you get all the current in the universe :) \$\endgroup\$ – pjc50 Aug 2 '13 at 17:11
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    \$\begingroup\$ Wow that must be a bright LED! \$\endgroup\$ – JYelton Aug 2 '13 at 17:29
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    \$\begingroup\$ @JYelton: For a tiny fraction of a second. Then it becomes the world's brightest arc lamp. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 2 '13 at 17:38
  • \$\begingroup\$ I wonder what is the current if you set the voltage to 30V. \$\endgroup\$ – Codism Aug 2 '13 at 19:18
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    \$\begingroup\$ @Codism just for fun I tried it, it stops the simulation after 6.1V, at which point the current is 464.16YA. lol. \$\endgroup\$ – MichaelK Aug 2 '13 at 20:30
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The simulator is pretending that the battery is an ideal power source, i.e. no internal resistance. As such, it is using the largest possible value it can support, 2147483647A.

True power sources do have an internal resistance, and this is what is preventing the LED you put across it from burning out. LEDs should have a resistor placed in series with them so that you don't have to depend on this internal resistance.

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    \$\begingroup\$ Ah, that makes sense. Thanks. (I would upvote and not comment but not enough rep...) \$\endgroup\$ – MichaelK Aug 2 '13 at 16:42
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    \$\begingroup\$ The simulator is also evidently failing to model LED bulk resistance. Even if the power source is idealized, there is no excuse for this. \$\endgroup\$ – Kaz Aug 2 '13 at 17:10
  • \$\begingroup\$ more on LED resistance: electronics.stackexchange.com/questions/76367/… \$\endgroup\$ – Phil Frost Aug 3 '13 at 1:19

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