I understand that each resistor in a circuit causes voltage drop. But according to Ohm’s law, the voltage drops to zero after the last resistor. So for the electrons that are traveling after the last resistor -where there is supposedly no voltage-, what’s pushing them to complete the remaining part of the circuit to get to the protons at the end.
It’s the electrons “behind” them that are pushing them along. When we say that the resistors have dropped the voltage to zero, we mean that a voltmeter measuring the voltage between the positive terminal and a nearby point on the wire would measure a voltage difference of almost zero (because the resistance in the wire itself is quite small). You could also have asked why current flows from one resistor to the “next” resistor, considering that the potential difference at two points on the wire that are between the same two resistors is nearly zero.
There is always a tiny portion of resistance in the wire and battery so there is always a few ele trons making the trip to the battery.
In tge ideal situation the resrance if tge wire is 0…
Then it does not need electrons to travel to the battery. The voltage at the wire end that has tge other end connected to the neg battery
A small part of the voltage drops in the wire resistance but it is so small, normally we ignore it. Your teacher is not going to mention these detail when they are trying to teach basic concepts. Voltage drop on wiring is a real problem, which is why power is distributed at high voltage and relatively lower current. If there is any current, then there is also a voltage since there is always some resistance.
EVERY conductor has resistance. It does take some energy to pass through that last wire from the end of the final resistor to the battery.
As wires are very good conductors this voltage is very low.
So we ignore it in most cases.
MOST cases but not all.
The neutral wire is connected to earth at the transformer. In my case this is some 500 m from my house.
When power is being drawn the “zero” end of the equipment is frequently running at 10 V or more.
This is the energy needed to force the current through that final wire back to the transformer.
As it is identical to the active wire on the way out that also loses the 10V so under load the volts available to me drop from around 245 down to perhaps 220.
Depending on the amount of current that I am using at the time.
The point is that the volts exist, they are both predictable and measurable but in many circuits the volts are sufficiently small that we don’t need to consider them.
The resistor will likely heat up to the point of burning up if you got one of high enough impedance to cause a voltage drop of zero so this question is not valid.
The voltage is not zero at the start of that last resistor. It’s only zero after passing all the way through that last one. So there is still voltage differential throughout that last resistor and that pushes the free electrons to to far side. But, a big BUT, those electrons are pushed on by the electrons flowing into them from upstream.
Remember, there is current flowing throughout the entire length of a closed circuit. In a series circuit for example, if there is a 1 amp flow out of the cathode, there is also a 1 amp flow into the anode.
The flow of electrons does not depend on the voltage. The voltage depends on Electron flow and resistance. If resistance is zero then voltage is zero, in a closed circuit.
A previous yahoo answer covered this subject on current and electrons. click link.
You have it backwards. Electrons flow opposite the direction of current. Since the conductor connects the resistor to the battery, electrons and current flow based upon Ohm’s Law. The only way to stop the current and electron flows is to open the circuit, by breaking it somewhere.
there is always a voltage drop, even in the best conductor
the resistance of the wire from the resistor to the battery may be fractions of an ohm, but it is still greater than zero