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When the AUX pin of a stereo is connected to my mobile phone's audio female port, I just measured the voltage at the end of this AUX pin(No load is connected) approximately 2mV using a DMM . But when I measured the current, it didn't indicate any current in the DMM. According to Ohm's law, voltage is current times the resistance. When relating this basic law to my practical case, it doesn't matching. How can it be realized?

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  • \$\begingroup\$ No, according to Ohm's law, current is actually voltage divided by resistance. No, according to Ohm's law, resistance is actually voltage divided by current! Which one is correct? A better question is: WHAT IS VOLTAGE?!!!! I mean, not the dry definition (frequently got wrong,) not equations. Write ten separate paragraphs each approaching voltage from a different aspect. Then you'll start to get this stuff. \$\endgroup\$ – wbeaty Sep 20 '18 at 3:14
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    \$\begingroup\$ You measured two different circuits, also you can't measure audio signal voltage reliably with a multimeter because it's not DC or sine-wave AC. \$\endgroup\$ – user253751 Sep 20 '18 at 3:33
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    \$\begingroup\$ Crossposted from physics.stackexchange.com/q/429763/2451 \$\endgroup\$ – Qmechanic Sep 20 '18 at 7:48
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I just measured the voltage at the end of this AUX pin(No load is connected) approximately 2mV using a DMM . But when I measured the current, it didn't indicate any current in the DMM.

By Ohm's Law

$$I = \frac{V}{R}$$

If you didn't connect any load, then \$R\approx\infty\$. Therefore, no matter what the voltage is you expect \$I\approx0\$.

In general, an ideal open circuit can sustain any voltage without passing any current. In practice, if there is air between the unconnected wire terminals, there is some limit to the voltage before a spark will be generated to carry charge between the terminals.

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