Nope... you'd probably read 0, because you'd be damaging stuff. On the off chance it survived you have two current sources in parallel not series, so you would only read 5 volts.Originally Posted by kryptkat
The one thing you do not do is back feed power into a current source... it lets all the smoke out and the part stops working.what about the circuitry ? voltage regulators ? or clipping diodes like the zener ? that would send any volts over 5v to ground leaving the 5v. so then you would only measure 5v ? assuming both power supplies have the same 5v regulator circuit.
And.. in this case the current source is the USB controller chip... not the power supply directly. The USB standard allows for 500ma (1/2 amp) per connector... 1/2 amp at 5 volts is 2.5 watts.
And what about that first instant when power is turned on but the capacitor is discharged, it's effective resistance is almost 0... there can be a huge burst of current... A 1uf electrolytic capacitor going abruptly from 0 to 10volts with a 1 ohm source resistance can pass a momentary burst of up to 10 amps!it is the current that blows parts. you can have a potential difference of charges extremely high as long as there is no current flow ie capacitor. same cap with a change of volts current will flow. ie sin or on off . the rating on the component says what it can take.
No it's a differential pair... when data+ is going positive data- is going negative at the same time. It's half-duplex communication.so the data + and data - are one way ? like full duplex ? the data + transmits while the data - receives ? or is it data + is a control line only two way and the data - is two way data only ?
