#### PConst167

##### Member

Hey friends,

this is a question that I've had for a long time. As we know, when a current passes through a resistor, a voltage drop occurs.

What I'd like to know is, what really is this voltage drop? Is it just the electrons bumping into the atomic structure of the resistor? There is something strange about this, because if voltage drop is just electrons bumping inside resistors, losing its energy, then why is it that the voltage drops always happen in exact ratios when you mix resistors in series? If electrons bumping inside a single resistor lose all their energy, then why don't they lose all their energy at the first resistor in a series of resistors?

It seems voltage "knows" how to divide perfectly between the resistors. So it must mean that voltage drop isn't simply electrons bumping against the atoms in a resistor. It seems more like a wave distributing among all the resistors at the same time. Let's say for example we have 2 resistors in series, separated by a distance D made of almost ideal wires. Could it actually be in fact, that after bumping inside the first resistor, the electrons regain their energy in the mean time inside the ideal wire, before entering the second resistor and losing that energy?

If so, then how is the energy gained exactly equal to the energy lost? How can voltage divide itself in perfect ratios ? How does it know!

If I put a multimeter across a resistor and measure the voltage, does the multimeter ever measure voltage itself? Or is it actually measuring the current and calculating the voltage via its internal resistor and using ohms law? Is it actually unaware of the voltage itself? Is it cheating!?

Since all voltages are really relative, then in fact all that must be happening is we are measuring currents and using resistors to find voltages through the abstract mathematical relation called Ohms law. So really the resistors and voltages don't know how to divide, it's just that we measure them in a way that makes it seem like they are.

So when measuring voltages across resistors in series, and the voltages seem to divide in perfect ratios, really what's happening is that by adding a second resistor, the current decreases, and then by attaching a large resistor in parallel with one of them will give a certain current in the large resistor, and then by Ohms law we calculate the voltages. And because Ohms law itself is a perfect mathematical ratio, and we use it, then we get perfect voltage ratios when measuring them.

What do you think?

this is a question that I've had for a long time. As we know, when a current passes through a resistor, a voltage drop occurs.

What I'd like to know is, what really is this voltage drop? Is it just the electrons bumping into the atomic structure of the resistor? There is something strange about this, because if voltage drop is just electrons bumping inside resistors, losing its energy, then why is it that the voltage drops always happen in exact ratios when you mix resistors in series? If electrons bumping inside a single resistor lose all their energy, then why don't they lose all their energy at the first resistor in a series of resistors?

It seems voltage "knows" how to divide perfectly between the resistors. So it must mean that voltage drop isn't simply electrons bumping against the atoms in a resistor. It seems more like a wave distributing among all the resistors at the same time. Let's say for example we have 2 resistors in series, separated by a distance D made of almost ideal wires. Could it actually be in fact, that after bumping inside the first resistor, the electrons regain their energy in the mean time inside the ideal wire, before entering the second resistor and losing that energy?

If so, then how is the energy gained exactly equal to the energy lost? How can voltage divide itself in perfect ratios ? How does it know!

If I put a multimeter across a resistor and measure the voltage, does the multimeter ever measure voltage itself? Or is it actually measuring the current and calculating the voltage via its internal resistor and using ohms law? Is it actually unaware of the voltage itself? Is it cheating!?

Since all voltages are really relative, then in fact all that must be happening is we are measuring currents and using resistors to find voltages through the abstract mathematical relation called Ohms law. So really the resistors and voltages don't know how to divide, it's just that we measure them in a way that makes it seem like they are.

So when measuring voltages across resistors in series, and the voltages seem to divide in perfect ratios, really what's happening is that by adding a second resistor, the current decreases, and then by attaching a large resistor in parallel with one of them will give a certain current in the large resistor, and then by Ohms law we calculate the voltages. And because Ohms law itself is a perfect mathematical ratio, and we use it, then we get perfect voltage ratios when measuring them.

What do you think?

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