I have a heating coil (actually an e-cigarette vaporizer) that I soldered two 10cm wires to. If I hook it up to a regulated 5V power source and measure the voltage over the coil, it measures about 4.07V. I find this puzzling because I can't figure out where the remaining 0.93 volts went. I'm measuring straight on the point where the wires connect to the power source leads. If I take away the coil, the leads measure exactly 5V.

I also have a 9V battery that, when measured, still has about 7.47V. If I connect the coil to that and, again, measure on the point where the wires connect to the battery leads, I measure only 2.39V over the coil, meaning 5,08V has somehow disappeared.

How is this happening? Is a heating element like this some sort of non-ohmic resistor? What exactly is going on here?

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    \$\begingroup\$ What is the actual part number of your "regulated 5 V power source"? Please provide a link to its datasheet if possible. How much current does your vaporizer draw? \$\endgroup\$ – The Photon Oct 30 '19 at 18:26
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    \$\begingroup\$ Bas, a 9V battery has a series resistance of something like 2 Ohms. That internal battery resistance will definitely lower the apparent voltage you see on the outside. \$\endgroup\$ – jonk Oct 30 '19 at 18:28
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    \$\begingroup\$ I'm with Jonk here. Your battery is losing steam because it's being pushed too hard. Internal series resistance of the battery. \$\endgroup\$ – DKNguyen Oct 30 '19 at 18:29
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    \$\begingroup\$ Possible duplicate of Why does measuring the voltage drop across a thing not simply measure the battery voltage? \$\endgroup\$ – Huisman Oct 30 '19 at 18:50
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    \$\begingroup\$ Odds are that you connected your wires in parallel, which halves the resistance and doubles the current, which quadruples the power required. \$\endgroup\$ – StainlessSteelRat Oct 30 '19 at 18:50

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