I Voltage drop: What do we mean when we say that voltage drops across a resistor?

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What do we mean when we say that voltage drops across a resistor?and what does it mean when we say that a point is at lower potential as compared to some other point?Can we say that potential is potential energy per unit charge?
 

phinds

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What do we mean when we say that voltage drops across a resistor?
We mean that the voltage is lower at one end of the resistor than it is at the other end.
and what does it mean when we say that a point is at lower potential as compared to some other point?
We mean that one point has a lower voltage than another point.
 

kuruman

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To add to @phinds's response
We mean that one end of the resistor is at a lower electrostatic potential than the other. Note that the convention in circuits is that straight lines are considered equipotentials. Here is a gravitational analogy to help you picture it. A straight line in a circuit is like walking a horizontal corridor - no change in gravitational potential energy. Stairs are like resistors. If you encounter stairs going up, you have a rise in gravitational potential from the bottom of the stairs to the top. If you encounter stairs going down, you have a drop in gravitational potential from the top of the stairs to the bottom. The same idea applies to electric potential. BTW, if you walk inside a building and visit various people on various floors but eventually end up where you started, the sum of all your potential rises and drops is zero. The same applies to charge carriers in electric circuits and the idea is called Kirchhoff's Voltage Law (KVL).
 
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To add to @phinds's response
We mean that one end of the resistor is at a lower electrostatic potential than the other. Note that the convention in circuits is that straight lines are considered equipotentials. Here is a gravitational analogy to help you picture it. A straight line in a circuit is like walking a horizontal corridor - no change in gravitational potential energy. Stairs are like resistors. If you encounter stairs going up, you have a rise in gravitational potential from the bottom of the stairs to the top. If you encounter stairs going down, you have a drop in gravitational potential from the top of the stairs to the bottom. The same idea applies to electric potential. BTW, if you walk inside a building and visit various people on various floors but eventually end up where you started, the sum of all your potential rises and drops is zero. The same applies to charge carriers in electric circuits and the idea is called Kirchhoff's Voltage Law (KVL).
Thanks now I am able to interpret. Can we say that after battery creates a potential diiference, electron gains potential energy as it passes through it and the difference in potential energy per unit charge between two pointsis called voltage between the two points. Am I right?
 
Thanks now I am able to interpret. Can we say that after battery creates a potential diiference, electron gains potential energy as it passes through it and the difference in potential energy per unit charge between two pointsis called voltage between the two points. Am I right?
Recall that the unit of voltage is the volt, which is equal to one joule per coulomb. So if you have a one volt battery, an electron passing through it will gain potential energy equal to the voltage of the battery times the charge of the particle. So yes, the particles passing through batteries will gain potential energy, but the magnitude of this energy gain is dependent on the charge of the particle.
 

vanhees71

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I'd say, a battery provides and electromotive force (emf) rather than a voltage in the strict sense. The word "force" is a bit unfortunate and must be read in the historical context. In the early 19th century, when the energy concept came up, famously mostly due to Helmholtz's discovery of what we call "energy conservation law" today, the word "force" meant in fact "energy". Helmholtz's famous paper is titled "Über die Erhaltung der Kraft" ("On the conservation of force"), but here "force" in fact means what we call "energy" today. Wikipedia gives a pretty good definition of the emf in its various forms

https://en.wikipedia.org/wiki/Electromotive_force
 

davenn

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Can we say that after battery creates a potential difference, the electron gains potential energy as it passes through it
No
An electron (electrons) don't pass through the battery. Very simply, a battery uses a chemical reaction
to separate charges. Negative charges ( electrons) move onto one terminal and positive charges ( ions)
onto the other terminal. This is what causes the potential difference ( the voltage measured on a voltmeter)
between the two terminals.
 
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I am a bit confused whether electron passes through a battery or no. My book says battery pumps electrons so it should be the case am I right?
 
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Can we say that potential difference is potential energy difference per unit charge. when positive charges move from higher potential to lower potential does their potential energy decrease . And when they they pass through battery does the potential energy per unit charge decrease.Can we conclude that potential at a point gives the potential energy per unit charge at that point or am I wrong please clarify
 

jbriggs444

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I am a bit confused whether electron passes through a battery or no. My book says battery pumps electrons so it should be the case am I right?
Forget you ever heard the word "electron". They are irrelevant to circuit theory.

In any case, no single electron needs to make a complete round trip through circuit and battery in order for current to flow.
 

vanhees71

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Hm, this is a bit problematic to say. To understand the constitutive laws, even in a schematic classical way, the idea of electrons are important.

E.g., often Ohm's law is stated as
$$\vec{j}=\sigma \vec{E},$$
without the qualification that this is valid in the non-relativistic limit only. Of course it's valid for usual household currents in usual household cables. How do you explain this, if not using electrons (though in the classical sense)?

The complete law (in Heaviside Lorentz units) must be
$$\vec{j}=\sigma(\vec{E}+\vec{v}/c \times \vec{B}).$$
This becomes important in the theory of the homopolar generator, which looks only "non-relativistic" on a superficial view. In this case the neglegence of relativity and (at least) classical electron theory leads to misconceptions and paradoxes!

Abandon the "fluid analogy" rather than the "classical-electron theory" as heuristics in circuit theory!
 

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