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Kirchhoff's law states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop”

What would happen if you placed a LED with a 2V drop into a circuit with a 12V battery and no resistor, would this work as it breaks Kirchhoff's law? and how would you calculate the current in this circuit?

Then, if you swapped the 12V battery with a 2V battery, what would happen as this is now in line with Kirchoff's law? what would the current be in this circuit?

Thanks.

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  • \$\begingroup\$ If you manage do create a circuit with really no resistance, please tell me \$\endgroup\$ – PlasmaHH Aug 12 '15 at 19:26
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    \$\begingroup\$ Well, even if you do, the voltage drop on the LED will be 12V. And it will blow up. 2V is a spec for a properly biased LED. \$\endgroup\$ – Eugene Sh. Aug 12 '15 at 19:29
  • \$\begingroup\$ @PlasmaHH We do it all the time. Weld some NbTi wire in a loop, chill, inject some current and off you go. \$\endgroup\$ – Spehro Pefhany Aug 12 '15 at 19:44
  • \$\begingroup\$ @SpehroPefhany I believe the LED won't be cool with the chilling.. \$\endgroup\$ – Eugene Sh. Aug 12 '15 at 19:45
  • \$\begingroup\$ @EugeneSh. They change color at 77K. Kind of fun. \$\endgroup\$ – Spehro Pefhany Aug 12 '15 at 19:46
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This would not break Kirchoff's law, but it would break the LED. Take a look at the current-voltage curve for a typical LED:

I-V for a typical LED

Here, around the forward operating voltage (Vd) the current (i) increases vastly with tiny changes in voltage. Vd in your case is 2V, typical for standard red, green and yellow LEDs. At 20V (not shown on the graph) the theoretical current would be phenomenal! This would be enough to vaporise most of the LED's constituent elements. In reality, what would happen is that the LED would shine very brightly for a very short time indeed before letting out its "magic smoke". This is the very reason that we include current limiting resistors in LED circuits.

If you swap the 12V battery for a 2V battery, things get interesting. Now we're right where the dotted line above Vd sits - small changes in voltage give large changes in current. Here, the current is defined by an exponential equation with a strong dependency on temperature. We won't know what the actual operating current is, but it will be strongly dependent on the exact battery voltage and the temperature, along with the battery's non-ideal series resistance (acting a bit like the series resistor we normally use with a diode). However no engineer worth their salt would ever drive a LED like this. LEDs are very sensitive to voltage changes, and their brightness (the parameter we want to control) is strongly related to their current. Hence LEDs are always driven with current in mind - be that with a statically chosen resistor or a specialised circuit with a feedback loop designed to keep current constant.

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  • \$\begingroup\$ Thanks @stefandz, so in a circuit with just an LED and 12V battery, 12V still flows throw the circuit (for a brief time at least), and the current would be dictated by the current-voltage curve? \$\endgroup\$ – RJSmith92 Aug 12 '15 at 20:27
  • \$\begingroup\$ The current voltage curve along with the internal resistance of the battery + wires. The two constraints would form a pair of simultaneous equations with one solution for current. \$\endgroup\$ – stefandz Aug 12 '15 at 20:28
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    \$\begingroup\$ @RJSmith92 Voltage doesn't flow, current does. If you blow up the LED, the voltage will still be there. \$\endgroup\$ – Mast Aug 12 '15 at 21:25
  • \$\begingroup\$ @Mast True, I should have worded it better. \$\endgroup\$ – RJSmith92 Aug 12 '15 at 22:02
  • \$\begingroup\$ @stefandz Thanks for the help with this. Final thing, the simple circuit from yesterday, with a 12V battery, 2V/20mA LED and 500 Ohm resistor, the current in the circuit was 20mA (I = 10/500 Ohm), so that's how the current was calculated. How does this tie in with the current-voltage curve? What's dictating the current in the circuit it? \$\endgroup\$ – RJSmith92 Aug 12 '15 at 22:15
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If you connect a standard LED across a 12 volt battery, a very large current (determined primarily by the wiring resistance) will flow briefly and the LED will vanish in a puff of smoke. KVL will be respected as long as the LED is intact.

The actual voltage across a lit LED is determined primarily by its composition (and therefore colour), and is highly unlikely to be 2.000 volts. The voltage will vary with current and temperature, so connecting a bare LED directly to a voltage source is a Bad Thing. LEDs must always be driven from a voltage source higher then their forward voltage, with a current limiting resistor or other current control device.

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A real circuit has non-idealities that would apply to this case:

  • An LED voltage is not exact: the current through a 2V LED will increase above the typical specified current as the voltage increases above 2V. 12V will cause the LED to fail.
  • All of the wires have some resistance: If the LED melts from passing too much power, it may become a very low resistance. Now the 12V supply may be driving perhaps 0.1 ohm resistance. If you have a 1500 Watt power that can drive 120 Amps and all of your wires are huge, this could theoretically be a stable state, but more likely parts of the circuit or power supply will start to melt until it is an open circuit. If the LED failure did not cause a short circuit but instead an open circuit, it is not as bad as there is no longer any closed circuit and no current will flow.
  • Also, the power supply has a maximum current: if you have a nice current-limited supply and you set the maximum current to 20mA, it will not actually be a 12V supply anymore, but more like a 2V supply. If the current limit is much higher or there is no current limit, see above.
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LED does have resistance, as well as the battery and almost anything else that can be connected to each other; no one knows "What would happen if you placed a LED with a 2V drop into a circuit with a 12V battery and no resistor".

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