Voltage drop in series [duplicate]

just wondering when there is a series circuit with one resistor that is supplied 60V, the voltage drop across that resistor is 60V.

However, if we add another resistor, the voltage drop across that same resistor from the example above becomes 30V.

Considering that the same total voltage is being applied to the same resistor, and the resistor is the first one to encounter the charge, why is the voltage drop different in the two situations? Ie, how does the circuit 'know' that there is another resistor that is coming up?

The resistors “know” each other through current. The current through a circuit depends on ALL the resistances along the path. So if you add a second equal resistor, the current will drop in half. And half the current through the same resistor is driven by half the voltage, which you can see from Ohm’s Law V=IR

• but looking at it in real life sort of way, why would the flow of charge from the starting point to the first resistor, or through the first resistor, be reduced simply because another resistor exists later along the path? It would be like saying, If I'm running through a forest, and I have to walk through a river, my speed is gonna slow down, but if I have to cross another river later on, I'm not gonna be aware of it and thus its not gonna affect my speed through the first river Commented Mar 7, 2023 at 13:56
• Think about it like you were in a line of people going through a tunnel. If the tunnel narrowed it would slow everyone even if they were at the back. Commented Mar 7, 2023 at 14:00

The total voltage from the source represents a fixed amount of electrical potential energy per unit charge (J/C) available to be "shared", if you will, by the resistors placed across the source in order to produce a given current (Coulombs per second, or C/s) through the total resistance and be dissipated as heat (J/s = J/C x C/s).

If it is one resistor, it gets all the voltage, and the current will equal the total voltage divided by that resistance. If it's two resistors, they are forced to "share" the total voltage in proportion to the magnitude of each resistance. The current will be the same total voltage but now divided by a greater resistance, making the current less than for the single resistor.

Each resistor doesn't have to "know" the other is present. There is only so much voltage (energy per unit charge) available to move the charge through the resistors.

Hope this helps.

Consider the water analogy. In the water analogy a voltage drop becomes a height drop. Water height represents potential energy, just like voltage. The flow rate is the current. Resistors become a restriction in flow, like a small pipe in a dam. In the picture below, you see two equal resistors in parallel. But the system is not in steady state yet: in steady state the voltage drops would be equal.

The flow rate through the pipes is given by $$I=\frac{U}{R}$$. The greater the drop in water height, the greater the pressure behind the pipe and so the greater the flow rate. In this case, since $$U_2>U_1$$, this means that $$I_2>I_1$$. Since the flow rate through the second dam is greater, the water level in the middle will drop until $$I_2=I_1$$. Since the resistors are equal here, the voltage drops will be the same: $$U_2=U_1$$.

For real electrical circuits the reasoning is the same, it just happens much quicker. If the currents didn't match, there would be charge building up and this is highly unstable.