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I purchased a Xiaomi laptop PRO computer, which uses a USB-C plug for its power input. I recently measured the plug's shell with a AC voltmeter, and found to my surprise a voltage between 90 to 100 volts (AC)! May be that's why I always felt the sting of a slight electrical shock every time I used the computer. I understand the shell is used for signal shielding and therefore should not be carrying any voltage. Why then is there a voltage? How can I get rid of it? Was this a design or manufacturing error? (The other end of the USB-C cable is connected to a 2 prong non-grounded 60W transformer plugged into the wall.)

ahchoooo
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    To what other point have you measured that voltage? If this weren't for safety reasons I'd vote to close this (the use of electronics is not on-topic) and your question is ill-formatted making it hard to read. My guess is that all is fine as your laptop would probably have some issues with AC in it. – Arsenal Apr 12 '18 at 15:19
  • What is the source impedance of this ~90VAC signal? – Voltage Spike Apr 12 '18 at 16:19
  • This is the usual thing, leakage. It is normally accepted and should be within regulation normative, see this topic as well, https://electronics.stackexchange.com/q/267146/117785 See also this, https://electronics.stackexchange.com/a/265375/117785 – Ale..chenski Apr 12 '18 at 17:48

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It's from an interference suppression capacitor between the shell and the mains AC. The capacitor provides a path for high frequency noise to be shunted to the power supply, but it also allows a tiny amount of current to pass through it. Because your multimeter is very high impedance, it registers as a high voltage, but if you connect it to something the voltage will drop a lot. Your observation of a slight electric shock fits with this as the capacitor won't allow enough current to provide more than a slight tingle.

C_Elegans
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