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I have a blue led light of 4Watt...one wire is connected to the AC source and the other one is not connected(1Live and 1 Neutral)...surprisingly enough the led lights up(very dim though)... how is it possible, we even dont have a closed circuit? Below are the imagesenter image description here

When One wire is connected--

When two wires are connected-- enter image description here

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    \$\begingroup\$ Probably the lamp is getting grounded by its surroundings, and has enough "ground" to power the circuitry. \$\endgroup\$ – Fusseldieb Sep 6 '17 at 19:57
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    \$\begingroup\$ The lamp is made out of which material (on the edges)? \$\endgroup\$ – Fusseldieb Sep 6 '17 at 19:59
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    \$\begingroup\$ Leakage current to ground via the fixture. \$\endgroup\$ – winny Sep 6 '17 at 20:03
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    \$\begingroup\$ There may be charged capacitors in the converter circuit somewhere slowly discharging through the LEDs? How long does the LED stay lit after turn-off? \$\endgroup\$ – calcium3000 Sep 6 '17 at 20:03
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    \$\begingroup\$ Does this happen when you interrupt the neutral and not the live? You should always interrupt the live. I noticed that many LED lamps (but also some compact fluorescent), tend to flicker if you interrupt the neutral wire but not the live. \$\endgroup\$ – next-hack Sep 6 '17 at 20:35
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This is power line electric field picked up open circuit wire. Proximity effects and stray capacitance close the loop to permit trace visible light on the order of xxuA is possible and is certainly visible. Normally common mode signals follow the LED wiring from any possible effect by electromagnetic coupling on 1 side and conductive coupling on the other.

The open circuit lightstrip acts as an antenna to AC electric field and rectifies via stray capacitance coupled E field also coupled as CM high frequency noise across ferrite transformers in SMPS.

The exact paths of signal coupling would require a full description of your components, layout and proximity of E fields etc.

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internal capacitors are charging on the positive half cycle of the AC wave, and discharging on the negative half. Very small amount of current. But Since leds will light up at very low currents, they turn on.

As mentioned in the comments, this will only happen if you only interrupt the neutral wire.

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