# Voltage drop remaining voltage [duplicate]

According to here the resistance drops liniar along the path.

In scenario 1 at the start of R1 the resistance the current encounters is 100$\Omega$ and at the end of R1 it encounters 0$\Omega$.

In scenario 2 at the start of R1 the resistance the current encounters is 200$\Omega$ and at the end of R1 it still encounters 100$\Omega$ because of R2.

If this is true, why does point A in scenario 2 have more volt left if it has encountered the higher resistance of 200$\Omega$ at the beginning opposed to scenario 1 where it encountered 100$\Omega$ at the beginning?

I know about Ohm's law and how to use it, but that doesn't explain how this occurres.

simulate this circuit – Schematic created using CircuitLab

Edit

I have written an answer here, thanks to answer from @Neil_UK in this post.

• Regardless of the total number of series of resistor and total value of these resistor, the voltage drop across the two end of battery must be 5V. Feb 10 '17 at 8:43
• Get a good physics book and read it. Or else get a small taste from here: electronics.stackexchange.com/a/272814/38098
– jonk
Feb 10 '17 at 8:45

The water analogy you linked to, pressure and flow down pipes, is very good up to a point, it is certainly good enough to handle cases like Ohms Law, and series resistors.

Perhaps it might be instructive to see what would happen if the voltage at point A in the second diagram was not 2.5v. Let's say that, to start with, we fix it at 0v by shorting R2 with a wire link, so we have something equivalent to the first diagram.

Now quickly remove the wire link. What happens?

With point A at 0V, there is 5v across R1, so by Ohms Law, 50mA flows through it. There is 0v across R2, so by Ohms Law, zero current flows through it. That means there's a current of 50mA flowing into point A.

Where does the current go? If this was a real experiment on the bench in front of you, the wires around point A would have some residual capacitance to ground, somewhere in the 1pF to 10pF region. If you had an oscilloscope probe or DVM monitoring point A, it would be more, 30pF for the scope, probably similar for the meter. The point is, there is some capacitance.

That current charges the capacitor, increasing its voltage. 50mA into 25pF causes it to charge up at 2GV/s, or 2v in 1nS, pretty quickly. So after 0.5nS, it will have got to around 1v.

Now what happens? R1 has 4v across it, so 40mA flows. R2 has 1v across it, so 10mA flows. Now only 30mA is flowing into point A, so the capacitor charges more slowly, but it's still charging.

It will keep charging until its voltage reaches 2.5v. At 2.5V, the current through both resistors is equal, and no current flows into point A, and the capacitance stops charging, the voltage stops changing. If, for some reason, the voltage rose above 2.5v (say you dropped the wire link onto R1 for a moment), then the capacitor would discharge back to 2.5v.

If you wanted to see that happening, then you could add a large capacitor between point A and 0v, to slow the whole process down so you could see it on a scope or a meter.

So you can see that the voltage on point A settles out, without the electricity having to 'know' anything about the circuit diagram, just responding according to Ohms Law and to the local resistor values.

What happens if there's no capacitance on point A? It's only in theory we can say it's zero, in simulators and in ideal mesh equations. In physical reality, there's always some present. But we can imagine what happens as we let the capacitance approach zero, the speed with which the node finds its 'correct' voltage approaches 'instantly'.