Mixing
![How can I get a sense of how management treats their staff before joining a company? [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjbpfN9tAutmK93bJRC3ZoROZzi2TJDms5n8_qJuhgE0a9b52OOHayv3NGT8igAdFL7byXNst-_1DZK5SjrIJ28_6RQPUpBROqMs5s6jo-ZsjX8kjDwfxJufIitH3TaQRXWaGSQKRQib-f/s72-c/1.jpg)
How do charges lose electrical potential energy when going through a resistor?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I can understand that the charges need to do work against the resistance - which transfers energy to forms such as light, heat, etcetera - using the electrostatic force provided by the battery terminals. But how does this result in a loss of electrical potential energy? The charges would only lose electrical potential energy if they were closer to the end terminal - in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor.
Help!
thermodynamics electric-circuits electric-current electrical-resistance dissipation
edited 47 mins ago
Qmechanic♦
98.2k121721066
asked 1 hour ago
Cr0xx
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
I can understand that the charges need to do work against the resistance - which transfers energy to forms such as light, heat, etcetera - using the electrostatic force provided by the battery terminals. But how does this result in a loss of electrical potential energy? The charges would only lose electrical potential energy if they were closer to the end terminal - in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor.
Help!
thermodynamics electric-circuits electric-current electrical-resistance dissipation
edited 47 mins ago
Qmechanic♦
98.2k121721066
asked 1 hour ago
Cr0xx
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I can understand that the charges need to do work against the resistance - which transfers energy to forms such as light, heat, etcetera - using the electrostatic force provided by the battery terminals. But how does this result in a loss of electrical potential energy? The charges would only lose electrical potential energy if they were closer to the end terminal - in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor.
Help!
thermodynamics electric-circuits electric-current electrical-resistance dissipation
edited 47 mins ago
Qmechanic♦
98.2k121721066
asked 1 hour ago
Cr0xx
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I can understand that the charges need to do work against the resistance - which transfers energy to forms such as light, heat, etcetera - using the electrostatic force provided by the battery terminals. But how does this result in a loss of electrical potential energy? The charges would only lose electrical potential energy if they were closer to the end terminal - in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor.
Help!
thermodynamics electric-circuits electric-current electrical-resistance dissipation
thermodynamics electric-circuits electric-current electrical-resistance dissipation
edited 47 mins ago
Qmechanic♦
98.2k121721066
asked 1 hour ago
Cr0xx
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 47 mins ago
Qmechanic♦
98.2k121721066
asked 1 hour ago
Cr0xx
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 47 mins ago
Qmechanic♦
98.2k121721066
edited 47 mins ago
Qmechanic♦
98.2k121721066
edited 47 mins ago
Qmechanic♦
98.2k121721066
98.2k121721066
asked 1 hour ago
Cr0xx
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Cr0xx
111
asked 1 hour ago
Cr0xx
111
111
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Cr0xx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
What you label as the "correct picture" - namely, the charge instantaneously losing potential as soon as it enters the resistor - is incorrect. Every resistive element in a circuit has a resistivity $rho$, which is an intrinsic property of the material. The resistance of a resistive element with length $L$ and (assumed uniform) cross-sectional area $A$ is
$$R=fracrho LA$$
In other words, $fracrhoA$ is the resistance per unit length of the resistor.
What this means is that our single resistor with resistance $R$ is equivalent to $N$ smaller resistors of length $fracLN$ and resistance $fracrho fracLNA=fracRN$. If we were to analyze this circuit of $N$ smaller resistors, we would find that there would be a voltage drop $V=IfracRN$ across each of them, for a given current $I$. This shows that the electrical potential energy of a charge depends on its position within the resistor. In fact, we can say even more than that: a charge that at position $x$ within the resistor (where $x=0$ is the entry point) has encountered a resistor of length $x$, and so the potential drop that that charge has experienced is, by Ohm's Law,
$$V(x)=Ifracrho xA$$
In other words, this shows that the potential difference increases linearly with position within the resistor (equivalently, the potential energy a charge has decreases linearly with distance). This energy loss is usually described as coming from collisions with the electrons and atoms within the resistor. Given that the collision rate with electrons and atoms in a resistor should be constant with time and position (as the intrinsic properties of the resistor don't change with position inside it), the answer we're getting makes physical sense.
In circuit analysis, typically we don't care about what's going on inside a resistor; rather, we're more interested in how the resistor as a whole behaves relative to other circuit elements. As such, we tend to use the lumped-element model in circuit analysis, where the effects of the resistor happen all in one "lump" that we place in a network of zero-resistance "wires." Since we don't tend to connect other circuit components to a point in the middle of a resistor (and if we do, we usually use a special variable-resistor lumped element), this model works quite well, as it only obscures the details that we don't care about in that situation. However, if you do care about what happens in the middle of a resistor (as you do in your question), then you can't use the lumped-element model to predict the physics, because it wasn't built to do that.
answered 39 mins ago
probably_someone
14.4k12451
add a comment |Â
up vote
1
down vote
in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor
Actually the boulder is rolling down a hill which has saplings growing on it.
The boulder accelerates (gains kinetic energy whilst losing gravitational potential energy) between hitting saplings.
When the boulder hits a sapling it loses some of its kinetic energy.
Thus the boulder rolls down the hill speeding up and then slowing down which when look on "from afar" it appears that the boulder is rolling down the hill at an approximately constant overall speed.
In statistics a Galton board is used as a demonstration to show the build up of a normal distribution but I want you to look at the passage of the balls down the board which replicate the boulder hitting saplings.
Now replace the word "gravitational" with "electric", "saplings" with "lattice ions" and "boulder" with "free electron" and you have a visualisation of the passage of an electron through a resistor.
answered 35 mins ago
Farcher
45.3k33388
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
add a comment |Â
up vote
0
down vote
Electromotive force is like a push required by electrons to get past the resistor. The loss of energy occurs because work has to be done against the resistance. If there was no resistor in the circuit, no EMF would be required. You can compare resistance to friction and EMF to the force applied in a object. In absence of any friction, the object will keep on moving indefinitely.
answered 57 mins ago
Mechanic7
303112
add a comment |Â
up vote
0
down vote
The charges lose potential energy by moving from a higher potential energy to a lower one. In the mean time they are accelerated and scattered. The result is an equilibrium drift speed, while the excess kinetic energy is transformed into heat.
answered 51 mins ago
my2cts
3,5922416
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
What you label as the "correct picture" - namely, the charge instantaneously losing potential as soon as it enters the resistor - is incorrect. Every resistive element in a circuit has a resistivity $rho$, which is an intrinsic property of the material. The resistance of a resistive element with length $L$ and (assumed uniform) cross-sectional area $A$ is
$$R=fracrho LA$$
In other words, $fracrhoA$ is the resistance per unit length of the resistor.
What this means is that our single resistor with resistance $R$ is equivalent to $N$ smaller resistors of length $fracLN$ and resistance $fracrho fracLNA=fracRN$. If we were to analyze this circuit of $N$ smaller resistors, we would find that there would be a voltage drop $V=IfracRN$ across each of them, for a given current $I$. This shows that the electrical potential energy of a charge depends on its position within the resistor. In fact, we can say even more than that: a charge that at position $x$ within the resistor (where $x=0$ is the entry point) has encountered a resistor of length $x$, and so the potential drop that that charge has experienced is, by Ohm's Law,
$$V(x)=Ifracrho xA$$
In other words, this shows that the potential difference increases linearly with position within the resistor (equivalently, the potential energy a charge has decreases linearly with distance). This energy loss is usually described as coming from collisions with the electrons and atoms within the resistor. Given that the collision rate with electrons and atoms in a resistor should be constant with time and position (as the intrinsic properties of the resistor don't change with position inside it), the answer we're getting makes physical sense.
In circuit analysis, typically we don't care about what's going on inside a resistor; rather, we're more interested in how the resistor as a whole behaves relative to other circuit elements. As such, we tend to use the lumped-element model in circuit analysis, where the effects of the resistor happen all in one "lump" that we place in a network of zero-resistance "wires." Since we don't tend to connect other circuit components to a point in the middle of a resistor (and if we do, we usually use a special variable-resistor lumped element), this model works quite well, as it only obscures the details that we don't care about in that situation. However, if you do care about what happens in the middle of a resistor (as you do in your question), then you can't use the lumped-element model to predict the physics, because it wasn't built to do that.
answered 39 mins ago
probably_someone
14.4k12451
add a comment |Â
up vote
1
down vote
What you label as the "correct picture" - namely, the charge instantaneously losing potential as soon as it enters the resistor - is incorrect. Every resistive element in a circuit has a resistivity $rho$, which is an intrinsic property of the material. The resistance of a resistive element with length $L$ and (assumed uniform) cross-sectional area $A$ is
$$R=fracrho LA$$
In other words, $fracrhoA$ is the resistance per unit length of the resistor.
What this means is that our single resistor with resistance $R$ is equivalent to $N$ smaller resistors of length $fracLN$ and resistance $fracrho fracLNA=fracRN$. If we were to analyze this circuit of $N$ smaller resistors, we would find that there would be a voltage drop $V=IfracRN$ across each of them, for a given current $I$. This shows that the electrical potential energy of a charge depends on its position within the resistor. In fact, we can say even more than that: a charge that at position $x$ within the resistor (where $x=0$ is the entry point) has encountered a resistor of length $x$, and so the potential drop that that charge has experienced is, by Ohm's Law,
$$V(x)=Ifracrho xA$$
In other words, this shows that the potential difference increases linearly with position within the resistor (equivalently, the potential energy a charge has decreases linearly with distance). This energy loss is usually described as coming from collisions with the electrons and atoms within the resistor. Given that the collision rate with electrons and atoms in a resistor should be constant with time and position (as the intrinsic properties of the resistor don't change with position inside it), the answer we're getting makes physical sense.
In circuit analysis, typically we don't care about what's going on inside a resistor; rather, we're more interested in how the resistor as a whole behaves relative to other circuit elements. As such, we tend to use the lumped-element model in circuit analysis, where the effects of the resistor happen all in one "lump" that we place in a network of zero-resistance "wires." Since we don't tend to connect other circuit components to a point in the middle of a resistor (and if we do, we usually use a special variable-resistor lumped element), this model works quite well, as it only obscures the details that we don't care about in that situation. However, if you do care about what happens in the middle of a resistor (as you do in your question), then you can't use the lumped-element model to predict the physics, because it wasn't built to do that.
answered 39 mins ago
probably_someone
14.4k12451
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What you label as the "correct picture" - namely, the charge instantaneously losing potential as soon as it enters the resistor - is incorrect. Every resistive element in a circuit has a resistivity $rho$, which is an intrinsic property of the material. The resistance of a resistive element with length $L$ and (assumed uniform) cross-sectional area $A$ is
$$R=fracrho LA$$
In other words, $fracrhoA$ is the resistance per unit length of the resistor.
What this means is that our single resistor with resistance $R$ is equivalent to $N$ smaller resistors of length $fracLN$ and resistance $fracrho fracLNA=fracRN$. If we were to analyze this circuit of $N$ smaller resistors, we would find that there would be a voltage drop $V=IfracRN$ across each of them, for a given current $I$. This shows that the electrical potential energy of a charge depends on its position within the resistor. In fact, we can say even more than that: a charge that at position $x$ within the resistor (where $x=0$ is the entry point) has encountered a resistor of length $x$, and so the potential drop that that charge has experienced is, by Ohm's Law,
$$V(x)=Ifracrho xA$$
In other words, this shows that the potential difference increases linearly with position within the resistor (equivalently, the potential energy a charge has decreases linearly with distance). This energy loss is usually described as coming from collisions with the electrons and atoms within the resistor. Given that the collision rate with electrons and atoms in a resistor should be constant with time and position (as the intrinsic properties of the resistor don't change with position inside it), the answer we're getting makes physical sense.
In circuit analysis, typically we don't care about what's going on inside a resistor; rather, we're more interested in how the resistor as a whole behaves relative to other circuit elements. As such, we tend to use the lumped-element model in circuit analysis, where the effects of the resistor happen all in one "lump" that we place in a network of zero-resistance "wires." Since we don't tend to connect other circuit components to a point in the middle of a resistor (and if we do, we usually use a special variable-resistor lumped element), this model works quite well, as it only obscures the details that we don't care about in that situation. However, if you do care about what happens in the middle of a resistor (as you do in your question), then you can't use the lumped-element model to predict the physics, because it wasn't built to do that.
answered 39 mins ago
probably_someone
14.4k12451
What you label as the "correct picture" - namely, the charge instantaneously losing potential as soon as it enters the resistor - is incorrect. Every resistive element in a circuit has a resistivity $rho$, which is an intrinsic property of the material. The resistance of a resistive element with length $L$ and (assumed uniform) cross-sectional area $A$ is
$$R=fracrho LA$$
In other words, $fracrhoA$ is the resistance per unit length of the resistor.
What this means is that our single resistor with resistance $R$ is equivalent to $N$ smaller resistors of length $fracLN$ and resistance $fracrho fracLNA=fracRN$. If we were to analyze this circuit of $N$ smaller resistors, we would find that there would be a voltage drop $V=IfracRN$ across each of them, for a given current $I$. This shows that the electrical potential energy of a charge depends on its position within the resistor. In fact, we can say even more than that: a charge that at position $x$ within the resistor (where $x=0$ is the entry point) has encountered a resistor of length $x$, and so the potential drop that that charge has experienced is, by Ohm's Law,
$$V(x)=Ifracrho xA$$
In other words, this shows that the potential difference increases linearly with position within the resistor (equivalently, the potential energy a charge has decreases linearly with distance). This energy loss is usually described as coming from collisions with the electrons and atoms within the resistor. Given that the collision rate with electrons and atoms in a resistor should be constant with time and position (as the intrinsic properties of the resistor don't change with position inside it), the answer we're getting makes physical sense.
In circuit analysis, typically we don't care about what's going on inside a resistor; rather, we're more interested in how the resistor as a whole behaves relative to other circuit elements. As such, we tend to use the lumped-element model in circuit analysis, where the effects of the resistor happen all in one "lump" that we place in a network of zero-resistance "wires." Since we don't tend to connect other circuit components to a point in the middle of a resistor (and if we do, we usually use a special variable-resistor lumped element), this model works quite well, as it only obscures the details that we don't care about in that situation. However, if you do care about what happens in the middle of a resistor (as you do in your question), then you can't use the lumped-element model to predict the physics, because it wasn't built to do that.
answered 39 mins ago
probably_someone
14.4k12451
answered 39 mins ago
probably_someone
14.4k12451
answered 39 mins ago
probably_someone
14.4k12451
answered 39 mins ago
probably_someone
14.4k12451
14.4k12451
add a comment |Â
add a comment |Â
up vote
1
down vote
in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor
Actually the boulder is rolling down a hill which has saplings growing on it.
The boulder accelerates (gains kinetic energy whilst losing gravitational potential energy) between hitting saplings.
When the boulder hits a sapling it loses some of its kinetic energy.
Thus the boulder rolls down the hill speeding up and then slowing down which when look on "from afar" it appears that the boulder is rolling down the hill at an approximately constant overall speed.
In statistics a Galton board is used as a demonstration to show the build up of a normal distribution but I want you to look at the passage of the balls down the board which replicate the boulder hitting saplings.
Now replace the word "gravitational" with "electric", "saplings" with "lattice ions" and "boulder" with "free electron" and you have a visualisation of the passage of an electron through a resistor.
answered 35 mins ago
Farcher
45.3k33388
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
add a comment |Â
up vote
1
down vote
in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor
Actually the boulder is rolling down a hill which has saplings growing on it.
The boulder accelerates (gains kinetic energy whilst losing gravitational potential energy) between hitting saplings.
When the boulder hits a sapling it loses some of its kinetic energy.
Thus the boulder rolls down the hill speeding up and then slowing down which when look on "from afar" it appears that the boulder is rolling down the hill at an approximately constant overall speed.
In statistics a Galton board is used as a demonstration to show the build up of a normal distribution but I want you to look at the passage of the balls down the board which replicate the boulder hitting saplings.
Now replace the word "gravitational" with "electric", "saplings" with "lattice ions" and "boulder" with "free electron" and you have a visualisation of the passage of an electron through a resistor.
answered 35 mins ago
Farcher
45.3k33388
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor
Actually the boulder is rolling down a hill which has saplings growing on it.
The boulder accelerates (gains kinetic energy whilst losing gravitational potential energy) between hitting saplings.
When the boulder hits a sapling it loses some of its kinetic energy.
Thus the boulder rolls down the hill speeding up and then slowing down which when look on "from afar" it appears that the boulder is rolling down the hill at an approximately constant overall speed.
In statistics a Galton board is used as a demonstration to show the build up of a normal distribution but I want you to look at the passage of the balls down the board which replicate the boulder hitting saplings.
Now replace the word "gravitational" with "electric", "saplings" with "lattice ions" and "boulder" with "free electron" and you have a visualisation of the passage of an electron through a resistor.
answered 35 mins ago
Farcher
45.3k33388
in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor
Actually the boulder is rolling down a hill which has saplings growing on it.
The boulder accelerates (gains kinetic energy whilst losing gravitational potential energy) between hitting saplings.
When the boulder hits a sapling it loses some of its kinetic energy.
Thus the boulder rolls down the hill speeding up and then slowing down which when look on "from afar" it appears that the boulder is rolling down the hill at an approximately constant overall speed.
In statistics a Galton board is used as a demonstration to show the build up of a normal distribution but I want you to look at the passage of the balls down the board which replicate the boulder hitting saplings.
Now replace the word "gravitational" with "electric", "saplings" with "lattice ions" and "boulder" with "free electron" and you have a visualisation of the passage of an electron through a resistor.
answered 35 mins ago
Farcher
45.3k33388
answered 35 mins ago
Farcher
45.3k33388
answered 35 mins ago
Farcher
45.3k33388
answered 35 mins ago
Farcher
45.3k33388
45.3k33388
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
add a comment |Â
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
Doesn't this analogy break down for wires with negligable resistance? Or are you saying that the resistance is the combination of the "hill" along with the "saplings"?
– Aaron Stevens
24 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
@AaronStevens The resistor is the hill and the saplings. For no resistance there is a smooth level ground and no saplings with the boulder rolling along at constant speed? Negligible resistance might be a very, very slight incline with very, very, few saplings?
– Farcher
6 mins ago
add a comment |Â
up vote
0
down vote
Electromotive force is like a push required by electrons to get past the resistor. The loss of energy occurs because work has to be done against the resistance. If there was no resistor in the circuit, no EMF would be required. You can compare resistance to friction and EMF to the force applied in a object. In absence of any friction, the object will keep on moving indefinitely.
answered 57 mins ago
Mechanic7
303112
add a comment |Â
up vote
0
down vote
Electromotive force is like a push required by electrons to get past the resistor. The loss of energy occurs because work has to be done against the resistance. If there was no resistor in the circuit, no EMF would be required. You can compare resistance to friction and EMF to the force applied in a object. In absence of any friction, the object will keep on moving indefinitely.
answered 57 mins ago
Mechanic7
303112
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Electromotive force is like a push required by electrons to get past the resistor. The loss of energy occurs because work has to be done against the resistance. If there was no resistor in the circuit, no EMF would be required. You can compare resistance to friction and EMF to the force applied in a object. In absence of any friction, the object will keep on moving indefinitely.
answered 57 mins ago
Mechanic7
303112
Electromotive force is like a push required by electrons to get past the resistor. The loss of energy occurs because work has to be done against the resistance. If there was no resistor in the circuit, no EMF would be required. You can compare resistance to friction and EMF to the force applied in a object. In absence of any friction, the object will keep on moving indefinitely.
answered 57 mins ago
Mechanic7
303112
answered 57 mins ago
Mechanic7
303112
answered 57 mins ago
Mechanic7
303112
answered 57 mins ago
Mechanic7
303112
303112
add a comment |Â
add a comment |Â
up vote
0
down vote
The charges lose potential energy by moving from a higher potential energy to a lower one. In the mean time they are accelerated and scattered. The result is an equilibrium drift speed, while the excess kinetic energy is transformed into heat.
answered 51 mins ago
my2cts
3,5922416
add a comment |Â
up vote
0
down vote
The charges lose potential energy by moving from a higher potential energy to a lower one. In the mean time they are accelerated and scattered. The result is an equilibrium drift speed, while the excess kinetic energy is transformed into heat.
answered 51 mins ago
my2cts
3,5922416
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The charges lose potential energy by moving from a higher potential energy to a lower one. In the mean time they are accelerated and scattered. The result is an equilibrium drift speed, while the excess kinetic energy is transformed into heat.
answered 51 mins ago
my2cts
3,5922416
The charges lose potential energy by moving from a higher potential energy to a lower one. In the mean time they are accelerated and scattered. The result is an equilibrium drift speed, while the excess kinetic energy is transformed into heat.
answered 51 mins ago
my2cts
3,5922416
answered 51 mins ago
my2cts
3,5922416
answered 51 mins ago
my2cts
3,5922416
answered 51 mins ago
my2cts
3,5922416
3,5922416
add a comment |Â
add a comment |Â
Cr0xx is a new contributor. Be nice, and check out our Code of Conduct.
Cr0xx is a new contributor. Be nice, and check out our Code of Conduct.
Cr0xx is a new contributor. Be nice, and check out our Code of Conduct.
Cr0xx is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f435717%2fhow-do-charges-lose-electrical-potential-energy-when-going-through-a-resistor%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
