In This Episode, Graham's Mind is Blown
This project got rid of all the extra stuff from last project, to focus on just the capacitor and resistors. There was a copious amount of reading, discussing what was actually going on in the resistor (both physically and mathematically) but the project was very simple.
Basically, I set the capacitor up in the circuit, put a voltage probe on it, and turn the circuit on. Watching the screen of the meter showed me the voltage increasing across the capacitor, as expected. Once again, I felt strongly that I really didn't understand what voltage is. Fortunately the book was about to discuss this in detail...
The other factor in this first part of the experiment was to have a resistor before the capacitor, and by swapping out a low-ohm resistor for a more resistive one, I could increase the time it took for the capacitor to charge. This was easily proved with the multimeter.
Then came all the theory. I learned about the way in which capacitors charge (they fill a fraction of the remaining difference in voltage over time, getting closer and closer to 'full' without every quite reaching it) which was apparently intuitive to me, The book also gave a formula for determining the charge time for a given capacitor and resistor pair.
Basically, I set the capacitor up in the circuit, put a voltage probe on it, and turn the circuit on. Watching the screen of the meter showed me the voltage increasing across the capacitor, as expected. Once again, I felt strongly that I really didn't understand what voltage is. Fortunately the book was about to discuss this in detail...
The other factor in this first part of the experiment was to have a resistor before the capacitor, and by swapping out a low-ohm resistor for a more resistive one, I could increase the time it took for the capacitor to charge. This was easily proved with the multimeter.
Then came all the theory. I learned about the way in which capacitors charge (they fill a fraction of the remaining difference in voltage over time, getting closer and closer to 'full' without every quite reaching it) which was apparently intuitive to me, The book also gave a formula for determining the charge time for a given capacitor and resistor pair.
There was a second part to the experiment as well: Put two resistors in series and measure the voltage across each.
It turns out that the resistance in a circuit always "contributes" 100% voltage drop. Again I don't really know what this means, but I can observe this: If I measure the voltage differential at two points on the same wire, even when 12v power is flowing through, I get 0 voltage measured on the meter. The wire doesn't resist so there's no differential between two points on it. If, however, I measure on each side of a resistor (any resistor!) I see 12 volts drop -- the entire voltage of the circuit.
(Get ready, Graham's mind is about to be blown.)
If you put two resistors in series, they each take a portion of the drop (such that both together drop a 12 volts -- the entire voltage of the circuit). If I put two identical resistors (say, 1k ohms each) then each one measures 6v, exactly half the drop. If I put in different resistor values, say 1k and 2k ohms, then the 1k resistor ends up dropping 4 volts, and the 2k resistor drops 8 volts.
It turns out that the resistance in a circuit always "contributes" 100% voltage drop. Again I don't really know what this means, but I can observe this: If I measure the voltage differential at two points on the same wire, even when 12v power is flowing through, I get 0 voltage measured on the meter. The wire doesn't resist so there's no differential between two points on it. If, however, I measure on each side of a resistor (any resistor!) I see 12 volts drop -- the entire voltage of the circuit.
(Get ready, Graham's mind is about to be blown.)
If you put two resistors in series, they each take a portion of the drop (such that both together drop a 12 volts -- the entire voltage of the circuit). If I put two identical resistors (say, 1k ohms each) then each one measures 6v, exactly half the drop. If I put in different resistor values, say 1k and 2k ohms, then the 1k resistor ends up dropping 4 volts, and the 2k resistor drops 8 volts.
Wait a second: my circuit can do math! BLAM!!!
By putting in different resistor ratios, I could read back those ratios of the original voltage on my meter. It makes perfect sense, that the ratio of the resistors would be the ratio of the voltage, but that means I could do something silly like, "What's two thirds of 12 volts?", put in a 1k and 2k resistor, measure across the 2k resistor, and pow! Multimeter tells me the answer is 8 volts!
I dunno. I was impressed.
Anyways, this basically explains why the capacitor charges the way it does: An uncharged capacitor has almost no resistance, and as it charges it's resistance increases to "infinite". The resistor chosen determines the voltage ratio in the early stages of charging, controlling how much apparent voltage the capacitor gets -- the more, the faster it charges. As the capacitor charges and it's resistance increases, it's owning more and more of the voltage drop and so less and less voltage is charging it.
Or something like that. I inferred all that from my measurements, and could have it all wrong! Clearly I need to do more research on the matter. This stuff is confusing!
And it looks like we're just starting to switch things up.
By putting in different resistor ratios, I could read back those ratios of the original voltage on my meter. It makes perfect sense, that the ratio of the resistors would be the ratio of the voltage, but that means I could do something silly like, "What's two thirds of 12 volts?", put in a 1k and 2k resistor, measure across the 2k resistor, and pow! Multimeter tells me the answer is 8 volts!
I dunno. I was impressed.
Anyways, this basically explains why the capacitor charges the way it does: An uncharged capacitor has almost no resistance, and as it charges it's resistance increases to "infinite". The resistor chosen determines the voltage ratio in the early stages of charging, controlling how much apparent voltage the capacitor gets -- the more, the faster it charges. As the capacitor charges and it's resistance increases, it's owning more and more of the voltage drop and so less and less voltage is charging it.
Or something like that. I inferred all that from my measurements, and could have it all wrong! Clearly I need to do more research on the matter. This stuff is confusing!
And it looks like we're just starting to switch things up.