Why didn't 0.2A at 2V kill me? [closed]

I was recently connecting a circuit together like here:

and I had the voltage set at 2V, with a 10Ω resistor. By Ohm's law, there was a current of 0.2A (and was confirmed by my multimeter). I accidentally touched both ends of the resistor...why wasn't I electrocuted? I thought the current necessary to kill a human was in the mA range.

closed as off-topic by Norbert Schuch, ACuriousMind♦, Gert, Bill N, JamalSFeb 14 '16 at 17:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Norbert Schuch, Gert, Bill N, JamalS
If this question can be reworded to fit the rules in the help center, please edit the question.

• Because your skin has a lot of resistance. – Mike Dunlavey Feb 13 '16 at 0:33
• Because you have a much higher resistance than the resistor, so no current flowed through you. – Jon Custer Feb 13 '16 at 0:33
• The current through the resistor is 0.2A but you're in parallel with the resistor, not in series. Parallel connected circuit elements have identical voltage across, not identical current through. – Alfred Centauri Feb 13 '16 at 0:34
• Really the question is if you thought that accidentally touching exposed wires in your experiment would kill you, why did you proceed with the experiment? – user10851 Feb 13 '16 at 0:35
• There are $2\,\text{Volts}$ across the resistor so according to $I = V/R$ there's $200\,\text{mA}$ going through it. Now you touch either side of the resistor. Your skin has, say, $10\,\text{k}\Omega$ resistance, so the $2\,\text{Volt}$ across your body produces $0.2\,\text{mA}$ current through you. There's no way $2\,\text{Volts}$ is ever going to kill you just from you touching it, because your skin has such a high resistance that such a low voltage puts a tiny current through you. The presence or absence of the resistor is completely irrelevant. – DanielSank Feb 13 '16 at 0:43

The power $P$ flowing through a circuit element with current $I$ undergoing voltage drop $V$ is given by $P=IV$. For your test circuit that is $\rm 0.2\,A \cdot 2\,V = 0.4\,W$, which is a quite modest power.
However, that's the power flowing through the resistor in the circuit that you drew. When you touched both leads of that resistor while the current was running, you added the impedance of your body in parallel with the resistance already in the circuit. A good ohmmeter will tell you that the resistance from one arm to the other is in the 100 k$\Omega$ range (but there's a lot of variation, mostly depending on whether your fingertips are sweaty or dry). A voltage drop of 2 V across a $10^5\,\Omega$ resistor gives a power of only $\rm 0.2\,nW$, entirely irrelevant to your body.