Electronics > Beginners
Some noob questions
Mr D:
Many thanks to all who are taking the time, i'm reading through all replies carefully and i believe it's having some positive effect!
How's this?:
Say i have a 3V battery in a simple circuit with only a switch (admittedly not a very sensible circuit, but for argument's sake).
Is it fair to say?: With the switch open, i'll measure around 3 volts across that open switch because (as there's no current flowing), that is the potential charge that could flow if i were foolish enough to close the switch?
So that's why if i remove the battery from the circuit and only measure across the terminals of the battery i'll measure 3 volts (ish), whereas if i short-circuit the terminals (not good, i know) and measure, i'll measure almost 0V as the maximum current possible is flowing at that moment so the potential current is almost zero?
Am i on the right track?
Old Printer:
I believe the reason your meter would read zero is because the dead short removes any difference between the prob tips, there is nothing to measure. Educated theory may offer another answer, but this is how I see it.
rstofer:
--- Quote from: Mr D on July 19, 2018, 12:57:21 pm ---Am i on the right track?
--- End quote ---
More or less... The short circuit of the switch will have a resistance of essentially (but not really) 0 Ohms. There will be no voltage drop across the switch because when you multiply by zero, the answer is 0. This isn't a special case, it follows from E = I * R. R = 0 so regardless of I, E will be 0. Ohm's LAW
Better, and safer, experiment: Connect the battery positive lead to your DMM red lead with the meter set to read milliamps. Connect the meter black lead to 2 each 1k resistors in series. You should read 3V / 2k = 1.5 mA. Now short one of the resistors. You should read 3V / 1k = 3 mA.
You can expand the experiment into a voltage divider circuit. You will probably want to measure voltage rather than current. Take out the meter, set it back to volts and reconnect the series resistors across the power supply. Clip the black lead to the wire headed to the negative battery terminal. You should read 1/2 the battery voltage at the center of the two series resistors.
Now change the top resistor to 100 Ohms. The reading should be 0.9 * the battery voltage. The calculation is easy: Battery voltage divided by total series resistance (to get the current through the loop) multiplied by the bottom resistor value (to get E = I * R for the bottom resistor and the loop current). Play with this until you really have a handle on the simple circuit. Swap the location of the 1k and 100 Ohm. The result would be 0.1 * the battery voltage.
The formula for the simple two resistor voltage divider is V = (E / (R1 + R2)) * R2 when R2 is the bottom resistor. The E / (R1 + R2) calculates the loop current I and then V = I * R2, the bottom resistor.
I like millamps, I don't like amps. Actually shorting a battery, particularly the more exotic batteries, will produce a LOT of current and it only takes a few seconds for the battery to catch fire. Then you have to run down the hall to the bathroom and flood the flaming mess with water. Just sayin'...
Brumby:
--- Quote from: Mr D on July 19, 2018, 12:57:21 pm ---So that's why if i remove the battery from the circuit and only measure across the terminals of the battery i'll measure 3 volts (ish), whereas if i short-circuit the terminals (not good, i know) and measure, i'll measure almost 0V as the maximum current possible is flowing at that moment so the potential current is almost zero?
Am i on the right track?
--- End quote ---
I'll say "Yes", with one qualification. The phrase "potential current" isn't correct. I would use the word "voltage" instead. Everything else reads like you are understanding correctly.
To understand where the voltage of the battery went, just look back at the internal resistance of the power source I mentioned earlier - ESR. The battery chemistry not only provides the voltage, but presents a certain amount of resistance as well - so the "equivalent circuit" of a battery is a voltage source and a resistor in series. As a result, when you short out a battery, the voltage generated within the battery is lost through the resistance of the battery.
james_s:
--- Quote from: rstofer on July 19, 2018, 02:43:16 pm ---I like millamps, I don't like amps. Actually shorting a battery, particularly the more exotic batteries, will produce a LOT of current and it only takes a few seconds for the battery to catch fire. Then you have to run down the hall to the bathroom and flood the flaming mess with water. Just sayin'...
--- End quote ---
Years ago I learned rather quickly that putting a 9V battery in your pocket with keys and loose change is a bad idea.
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