Posted: 4/24/2005 10:05:25 PM EDT
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Maybe I'm doing something wrong here but I feel like a complete retard. Last week I asked which resistor would take 5v down to 3.4v. We came to the conclusion that something in the 60ohm range would take care of voltage. So I go buy some resistors and start playing around. I took the 5v + feed and soldered the resistor to it. I then turned on the power supply and took a voltage reading putting my + lead on the other side of the resistor and the - lead on the 5v - feed. I was suprised to see that the voltage was exactly the same on both sides of the resistor. I thought it should have givin me a reduced reading. So I hooked up a 150ohm resistor exactly the same way and once again the voltage was the same on both sides of the resistor. What am I doing wrong here? Do I just not understand how this should work? Any info would be greatly appreciated! |
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There is no voltage drop because there is no load between ground and the other lead of the resistor. Put the LED or a resistor in the circuit, you will measure a lower voltage at that point. V+ (+5v) -----/\/\/\/\/-----(+5v) still 5 v as the current ain't going nowhere! There is no voltage drop because it has no where to drop to. Duh! As ColKlink pointed out, V=IR. Since the resistor is not connected to anything, there is no current. V = 0 x R = 0. Zero voltage drop. V+ (+5v) ----/\/\/\/\/------ V- or ground the voltage will measure zero at this point. V+ (+5v) -----/\/\/\/\/------ Vx -------/\/\/\/\/------V- or ground the voltage at Vx will measure somewhere between +5v and 0v depending on the value of the two resistors. V+ (+5v) ------/\/\/\/\-------(>|)----- V- or ground. |
I appreciate the information, but DUh? Where do you come off. I take it your were born with that information and didnt learn it from someone else? |