0
$\begingroup$

I feel that I don't understand how voltage works in a circuit. I understand voltage to be electric potential energy per unit charge $$kq/r.$$ In the case of a circuit, electrons flow from low potential to high potential. But I don't understand how resistors cause a voltage drop. Isn't voltage based on position? How can the resistors cause a drop in potential energy? I can understand the resistor causing a drop in kinetic energy slowing electrons down, but how does it lower voltage? And if the total voltage is equal to the sum of the voltages of the resistors, then the voltage difference in the wire after the last resistor and the positive terminal would be zero, right? Then how would the electrons be able to flow back to the positive terminal? Wouldn't they just stop?

The best analogy I can think of is a river flowing downhill that turns a turbine. But in this case the turbine doesn't cause a drop in the water's potential energy. It only takes some of the water's kinetic energy.

$\endgroup$
  • $\begingroup$ In the case of a turbine, water has a potential energy which gets converted to kinetic when it flows downstream. So, indirectly, its the potential energy getting converted to kinetic. $\endgroup$ – Shubham Oct 5 '15 at 17:55
  • $\begingroup$ Right but what about the circuit case..can you explain? $\endgroup$ – physm12 Oct 6 '15 at 13:45
  • $\begingroup$ The issue I am having is that voltage is treated a total energy per unit charge but in electrostatics it was defined as POTENTIAL energy per unit charge. In a circuit, I understand across a wire with zero resistance there should be negligible energy loss. But shouldn't there still be a voltage drop. Wouldn't potential energy be converted to kinetic energy? $\endgroup$ – physm12 Oct 10 '15 at 10:44
1
$\begingroup$

Isn't voltage based on position?

No. Any point in a conducting wire ie has the same potential (ideally).

How can the resistors cause a drop in potential energy? I can understand the resistor causing a drop in kinetic energy slowing electrons down, but how does it lower voltage?

If potential is analogous to pressure (and voltage thus to pressure difference), then in order to move for example water through a filter you must add pressure from one side. Not from the other. There is a pressure difference across the filter. Similarly there is a voltage drop across a resistor because a larger potential is needed to push charges through from one side.

the voltage difference in the wire after the last resistor and the positive terminal would be zero, right?

Yes. Again, because the potential at any point on a conducting wire is the same.

By the way, avoid the word "voltage difference". "Voltage" already means "potential difference".

Then how would the electrons be able to flow back to the positive terminal? Wouldn't they just stop?

Why would they stop? What stops them? If the wire is indeed one hundred percent conducting (ideally has no resistance), then they just continue unhindered. Just like a spaceship continues drifting if nothing stops it.

Don't confuse motion with acceleration. To start or stop something we need a force. An electric force is required to overcome the resistance force in the resistor, otherwise the charges would stop there. But there is no such resisting force in a wire.

The best analogy I can think of is a river flowing downhill that turns a turbine. But in this case the turbine doesn't cause a drop in the water's potential energy. It only takes some of the water's kinetic energy.

It is a fine analogy, just keep in mind that gravitational potential energy is something different than electrical potential energy. There is potential energy present both in water-flowing-down-a-stream and in electrons-moving-through-a-circuit.

The electric potential is not analogous to the gravitational potential energy in the water scenario, but rather to the pressure. Because they work in the same way (difference in pressure means that something is pushed more form one side, so it will want to move - electric potential difference similarly means that charge is being repelled more from one side than the other, so it will want to move.

$\endgroup$
0
$\begingroup$

Let's see...let's start with Ohm's law: V=IxR
We'll now refer to a series circuit Any resistor (or component) will have a voltage drop across it,given by the relationship: Vdrop=IxR(R is the resistance of that component).Note that intensity will be the same in any point,so if it will decrease,it will do it globaly(total intensity).Voltage may decrease locally(just in a certain point) or the total voltage of the circuit will decrease.This is because of the way component resistance behaves.

And if the total voltage is equal to the sum of the voltages of the resistors, then the voltage difference in wire after the last resistor and the positive terminal would be zero, right? Then how would the electrons be able to flow back to the positive terminal? Wouldn't they just stop?

Actually,the wire also has a voltage drop.It also has it when the voltage source is short circuited,too,but it's too small and it allows to much current to pass through.

$\endgroup$
0
$\begingroup$

Short answer:

The resistor does not actually cause a "voltage drop" -- at least not directly. The potential difference across the resistor is better understood as a natural consequence of the fact that nonzero electromagnetic fields must be present within the resistor in order for a current to flow through it.

Long answer:

Consider a charge-neutral metal conductor that contains free electrons, electrons that are not bound to the atoms in the metal lattice and are therefore allowed to move freely around the material. Let the lengthwise dimension of the conductor be denoted as $L$ and the cross-section area be denoted as $A$, and for simplicity, let both of these parameters be constant. In equilibrium, the free electrons experience thermal motion which causes them to collide with the atomic lattice that makes up the bulk of the material, and the overall thermal motion of the free electrons is random such that there is no bias for a given electron to move in a specific direction. Thus the average velocity of any given free electron is $0$.

However, applying a constant $\vec E$-field in the direction parallel to $L$ will introduce a force that accelerates the electrons in the direction opposite of $\vec E$. Now that the electrons are in a non-equilibrium state, the rate of acceleration for the electrons will decrease due to collisions with the atomic lattice and other thermal effects until an equilibrium state is achieved in which the average velocity of any given electron is constant and non-zero in the direction opposite of $\vec E$. This average velocity is referred as the drift velocity of the free electrons in the metal, and the drift of these electrons is what is responsible for the macroscopic observation of a positive current $\vec I$ flowing in the direction of $\vec E$.

What I have just described is the conceptual framework for what is known as the Drude model, a classical model of electrical conduction in metals and other conductive materials based on the principles of classical electromagnetism and kinetic theory. There are more generalized and modern variations of this model that incorporate our modern understanding of electromagnetic phenomena, but this model turns out to be sufficient for giving theoretical justification for Ohm's Law, which in vector form is:

$$\vec E = \rho \vec J$$

where $\vec E$ is the applied electric field within the conductor, $\vec J = \frac{\vec I}{A}$ is the current density (the amount of current per area flowing through the cross-section area of the conductor), and $\rho$ is the resistivity of the conductor, an intrinsic property of the material.

Now consider one of the free electrons moving along the length of the metal. The amount of work per unit charge $W$ performed by the $\vec E$-field on the electron is defined as the line integral of $\vec E$ along the path $C$ that the electron takes through the metal,

$$W = \int_C \vec E \cdot d \vec l$$

Putting together this expression for work and Ohm's Law, we end up with

$$\int_C \vec E \cdot d \vec l = \int_C \rho \frac{\vec I}{A} \cdot d \vec l$$

But the left-hand side of this equation is exactly the definition of an electromotive force $V_{EMF}$ -- the amount of work per unit charge performed by the $\vec E$-field along the path that the electron takes through the metal -- and by assuming the electron moves with constant drift velocity $\vec v_d$, the path $C$ the electron takes through the metal can be assumed to be a straight line with length L, and the right-hand side of the equation reduces to

$$\int_C \rho \frac{\vec I}{A} \cdot d \vec l = \rho \frac{I}{A} L$$

Thus,

$$V_{EMF} = IR$$

where

$$R \equiv \frac{\rho L}{A}$$

is the resistance of the metal conductor with length $L$ and cross-section area $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.