I would assume that the poor resistor was smoking in hell because P=VI (20Volts x 0.2A = 4Watts of power). And I would also assume that besides getting me the wrong resistor (100 Ohms instead of 1K), the resistor my lab partner gave me was rated at maybe 1/8 Watt. 1/8 watt max receiving 4 watts of power is like he-man power flowing through a beggar that hasn't eaten for a month! He melted 🤣😂🙃
At 7:07 you say "what goes up must come down". And somehow this means that after the first little resistive load the voltage drops to zero? I'm pretty sure there is still some voltage to push current through a second small resistive load you might add in series. And the second load wouldn't magically make the voltage drop to zero, either. Whether your meter probes contact the source terminals directly or at nodes A and B, you are measuring the same source voltage! It is not at all clear to me what the concept is supposed to be here or how it is supposed to work. It is clear that you are talking about *something important* that I need to understand, but it can't be actual voltage. Even an open circuit instead of a resistive load would still make the voltage read the same as the source voltage, measured in the way that you show in the diagram. As implied by Ohm's law, the same voltage applied to more resistance yields less current. Not simply zero voltage, which *seems* to be what you are saying, but can't be what you actually mean. Any clarity on what is actually meant here would be very much appreciated. Moreover, does conventional current travel away from positive terminals and toward negative terminals? The answer here seems to be that it changes directions. But of course it doesn't. It consistently flows in the same direction around the entire circuit. So why on earth are we calling the positive terminal from the electrical source the terminal from which current flows, but the negative terminal of a resistive load the terminal from which current flows? Since the direction of flow isn't changing, why is the convention about current flow inconsistent with the actual direction of flow here?
Lots to unpack here. First, yes conventional current travels from the + terminal of a source to the - terminal of a source. This is NOT how real electrons flow. Electrons actually flow from negative to positive. It doesn't matter which direction they travel as long as there is a flow through something. DC is one direction only. AC alternates back and forth. The difference between conventional and electron flow only becomes more important when we discuss the inner workings of semiconductor devices and circuits. In this simplification we're assuming the wires have 0 resistance. In a real circuit yes there would be a tiny, tiny, tiny amount of residual voltage necessary to continue to push the current back to the source negative terminal to account for the tiny, tiny, tiny resistance of the wire.
Definitely will be revisiting this lecture over and over again. Thanks for the info.
Make sure you put yourself to the test! Check out the examples lecture at: th-cam.com/video/tGfbE61A_mU/w-d-xo.html
@@bigbadtech Appreciated
Love your work, Jim!
All you need now is a UniBrow and some minor explosions and you will have 5 million subscribers. ;-)
(You got one here for sure)
I'm definitely not as "flashy" as Mehdi!
I would assume that the poor resistor was smoking in hell because P=VI (20Volts x 0.2A = 4Watts of power). And I would also assume that besides getting me the wrong resistor (100 Ohms instead of 1K), the resistor my lab partner gave me was rated at maybe 1/8 Watt. 1/8 watt max receiving 4 watts of power is like he-man power flowing through a beggar that hasn't eaten for a month! He melted 🤣😂🙃
At 7:07 you say "what goes up must come down". And somehow this means that after the first little resistive load the voltage drops to zero? I'm pretty sure there is still some voltage to push current through a second small resistive load you might add in series. And the second load wouldn't magically make the voltage drop to zero, either. Whether your meter probes contact the source terminals directly or at nodes A and B, you are measuring the same source voltage! It is not at all clear to me what the concept is supposed to be here or how it is supposed to work. It is clear that you are talking about *something important* that I need to understand, but it can't be actual voltage. Even an open circuit instead of a resistive load would still make the voltage read the same as the source voltage, measured in the way that you show in the diagram. As implied by Ohm's law, the same voltage applied to more resistance yields less current. Not simply zero voltage, which *seems* to be what you are saying, but can't be what you actually mean. Any clarity on what is actually meant here would be very much appreciated.
Moreover, does conventional current travel away from positive terminals and toward negative terminals? The answer here seems to be that it changes directions. But of course it doesn't. It consistently flows in the same direction around the entire circuit. So why on earth are we calling the positive terminal from the electrical source the terminal from which current flows, but the negative terminal of a resistive load the terminal from which current flows? Since the direction of flow isn't changing, why is the convention about current flow inconsistent with the actual direction of flow here?
Lots to unpack here. First, yes conventional current travels from the + terminal of a source to the - terminal of a source. This is NOT how real electrons flow. Electrons actually flow from negative to positive. It doesn't matter which direction they travel as long as there is a flow through something. DC is one direction only. AC alternates back and forth. The difference between conventional and electron flow only becomes more important when we discuss the inner workings of semiconductor devices and circuits.
In this simplification we're assuming the wires have 0 resistance. In a real circuit yes there would be a tiny, tiny, tiny amount of residual voltage necessary to continue to push the current back to the source negative terminal to account for the tiny, tiny, tiny resistance of the wire.
Absolutely so annoying
Absolutely.