Capacitor and inductor working principle
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I want to know that there is opposition mechanism of current in inductor say back emf than what is opposition mechanism of voltage in capacitor ??? And also want to know that when inductor suddenly open it generates very huge voltage to keep current constant does it violate conservation of energy ???
power-electronics power-engineering electrical engineering
asked Aug 8 at 12:32
user48391
161
add a comment |Â
up vote
3
down vote
favorite
I want to know that there is opposition mechanism of current in inductor say back emf than what is opposition mechanism of voltage in capacitor ??? And also want to know that when inductor suddenly open it generates very huge voltage to keep current constant does it violate conservation of energy ???
power-electronics power-engineering electrical engineering
asked Aug 8 at 12:32
user48391
161
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I want to know that there is opposition mechanism of current in inductor say back emf than what is opposition mechanism of voltage in capacitor ??? And also want to know that when inductor suddenly open it generates very huge voltage to keep current constant does it violate conservation of energy ???
power-electronics power-engineering electrical engineering
asked Aug 8 at 12:32
user48391
161
I want to know that there is opposition mechanism of current in inductor say back emf than what is opposition mechanism of voltage in capacitor ??? And also want to know that when inductor suddenly open it generates very huge voltage to keep current constant does it violate conservation of energy ???
power-electronics power-engineering electrical engineering
power-electronics power-engineering electrical engineering
asked Aug 8 at 12:32
user48391
161
asked Aug 8 at 12:32
user48391
161
asked Aug 8 at 12:32
user48391
161
asked Aug 8 at 12:32
user48391
161
asked Aug 8 at 12:32
user48391
161
161
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
It's very fruitful to compare inductors and capacitors. They are dual components. They both store energy. However, you have to remember to swap current and voltage, short circuit and open circuit, series and parallel.
A capacitor 'tries to' keep its terminal voltage constant. If you try to change the voltage suddenly, it will demand a large current. 'Nothing' happens when the capacitor is open circuit, it just stays at the same voltage. You can ramp the charge stored in it by supplying a current.
An inductor 'tries to' keep its current constant. If you try to change the current suddenly, it will generate a large voltage. 'Nothing' happens when the inductor is short-circuited, it just stays at the same current. You can ramp the current through it by supplying a voltage.
The main differences are that people tend to have a better feel for capacitors, voltage sources feel more natural than current sources.
Capacitors also tend to be more ideal, typical capacitors have losses that are far smaller than those of typical inductors. An open circuit capacitor can store charge for seconds or minutes, there's time to measure the voltage with a DMM. A typical inductor shorted through a current meter will drop the current to zero too fast to be seen with teh naked eye. This difference is not fundamental to inductors and capacitors, but is just the way the properties of copper and plastic happen to turn out. If we had room temperature superconductors, and only damp cardboard to insulate capacitors, then inductors would be the ideal components.
answered Aug 8 at 12:55
Neil_UK
70.6k273155
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
add a comment |Â
up vote
3
down vote
For an inductor: -
$$V = Ldfracdidt$$
This means a back emf voltage is produced when the current tries to change - that back emf tries to oppose the change in current.
For a capacitor: -
$$I = Cdfracdvdt$$
This means that a current is produced when the voltage tries to change and that current tries to oppose the change in voltage.
And also want to know that when inductor suddenly open it generates
very huge voltage to keep current constant does it violate
conservation of energy ?
No, there is no violation of energy conservation. The amount of energy an inductor can use to produce a high back emf is limited by the current passing through the inductor when the terminals were open circuited. That energy is: -
$$W = dfrac12cdot LI^2$$
So, if the current was 10 amps when a 100 uH inductor open-circuited, the energy of the spark can be no more than 5 milli joules.
answered Aug 8 at 12:53
Andy aka
232k10172395
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
1
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
It's very fruitful to compare inductors and capacitors. They are dual components. They both store energy. However, you have to remember to swap current and voltage, short circuit and open circuit, series and parallel.
A capacitor 'tries to' keep its terminal voltage constant. If you try to change the voltage suddenly, it will demand a large current. 'Nothing' happens when the capacitor is open circuit, it just stays at the same voltage. You can ramp the charge stored in it by supplying a current.
An inductor 'tries to' keep its current constant. If you try to change the current suddenly, it will generate a large voltage. 'Nothing' happens when the inductor is short-circuited, it just stays at the same current. You can ramp the current through it by supplying a voltage.
The main differences are that people tend to have a better feel for capacitors, voltage sources feel more natural than current sources.
Capacitors also tend to be more ideal, typical capacitors have losses that are far smaller than those of typical inductors. An open circuit capacitor can store charge for seconds or minutes, there's time to measure the voltage with a DMM. A typical inductor shorted through a current meter will drop the current to zero too fast to be seen with teh naked eye. This difference is not fundamental to inductors and capacitors, but is just the way the properties of copper and plastic happen to turn out. If we had room temperature superconductors, and only damp cardboard to insulate capacitors, then inductors would be the ideal components.
answered Aug 8 at 12:55
Neil_UK
70.6k273155
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
add a comment |Â
up vote
7
down vote
It's very fruitful to compare inductors and capacitors. They are dual components. They both store energy. However, you have to remember to swap current and voltage, short circuit and open circuit, series and parallel.
A capacitor 'tries to' keep its terminal voltage constant. If you try to change the voltage suddenly, it will demand a large current. 'Nothing' happens when the capacitor is open circuit, it just stays at the same voltage. You can ramp the charge stored in it by supplying a current.
An inductor 'tries to' keep its current constant. If you try to change the current suddenly, it will generate a large voltage. 'Nothing' happens when the inductor is short-circuited, it just stays at the same current. You can ramp the current through it by supplying a voltage.
The main differences are that people tend to have a better feel for capacitors, voltage sources feel more natural than current sources.
Capacitors also tend to be more ideal, typical capacitors have losses that are far smaller than those of typical inductors. An open circuit capacitor can store charge for seconds or minutes, there's time to measure the voltage with a DMM. A typical inductor shorted through a current meter will drop the current to zero too fast to be seen with teh naked eye. This difference is not fundamental to inductors and capacitors, but is just the way the properties of copper and plastic happen to turn out. If we had room temperature superconductors, and only damp cardboard to insulate capacitors, then inductors would be the ideal components.
answered Aug 8 at 12:55
Neil_UK
70.6k273155
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
add a comment |Â
up vote
7
down vote
up vote
7
down vote
It's very fruitful to compare inductors and capacitors. They are dual components. They both store energy. However, you have to remember to swap current and voltage, short circuit and open circuit, series and parallel.
A capacitor 'tries to' keep its terminal voltage constant. If you try to change the voltage suddenly, it will demand a large current. 'Nothing' happens when the capacitor is open circuit, it just stays at the same voltage. You can ramp the charge stored in it by supplying a current.
An inductor 'tries to' keep its current constant. If you try to change the current suddenly, it will generate a large voltage. 'Nothing' happens when the inductor is short-circuited, it just stays at the same current. You can ramp the current through it by supplying a voltage.
The main differences are that people tend to have a better feel for capacitors, voltage sources feel more natural than current sources.
Capacitors also tend to be more ideal, typical capacitors have losses that are far smaller than those of typical inductors. An open circuit capacitor can store charge for seconds or minutes, there's time to measure the voltage with a DMM. A typical inductor shorted through a current meter will drop the current to zero too fast to be seen with teh naked eye. This difference is not fundamental to inductors and capacitors, but is just the way the properties of copper and plastic happen to turn out. If we had room temperature superconductors, and only damp cardboard to insulate capacitors, then inductors would be the ideal components.
answered Aug 8 at 12:55
Neil_UK
70.6k273155
It's very fruitful to compare inductors and capacitors. They are dual components. They both store energy. However, you have to remember to swap current and voltage, short circuit and open circuit, series and parallel.
A capacitor 'tries to' keep its terminal voltage constant. If you try to change the voltage suddenly, it will demand a large current. 'Nothing' happens when the capacitor is open circuit, it just stays at the same voltage. You can ramp the charge stored in it by supplying a current.
An inductor 'tries to' keep its current constant. If you try to change the current suddenly, it will generate a large voltage. 'Nothing' happens when the inductor is short-circuited, it just stays at the same current. You can ramp the current through it by supplying a voltage.
The main differences are that people tend to have a better feel for capacitors, voltage sources feel more natural than current sources.
Capacitors also tend to be more ideal, typical capacitors have losses that are far smaller than those of typical inductors. An open circuit capacitor can store charge for seconds or minutes, there's time to measure the voltage with a DMM. A typical inductor shorted through a current meter will drop the current to zero too fast to be seen with teh naked eye. This difference is not fundamental to inductors and capacitors, but is just the way the properties of copper and plastic happen to turn out. If we had room temperature superconductors, and only damp cardboard to insulate capacitors, then inductors would be the ideal components.
answered Aug 8 at 12:55
Neil_UK
70.6k273155
edited Aug 8 at 14:07
answered Aug 8 at 12:55
Neil_UK
70.6k273155
answered Aug 8 at 12:55
Neil_UK
70.6k273155
answered Aug 8 at 12:55
Neil_UK
70.6k273155
70.6k273155
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
add a comment |Â
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
The other big difference is that at least at room temperature real capacitors have a much larger "self time constant" than real inductors. So we can do experiments with capacitors on "human" timescales while it is much harder to do that with inductors.
– Peter Green
Aug 8 at 15:40
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
@PeterGreen I think I was saying that when I said you had time to measure a capacitor with a DMM, but inductor current dropped too fast to see.
– Neil_UK
Aug 8 at 18:56
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
Brainfart, seems I missed the last paragraph of your answer.
– Peter Green
Aug 8 at 19:00
add a comment |Â
up vote
3
down vote
For an inductor: -
$$V = Ldfracdidt$$
This means a back emf voltage is produced when the current tries to change - that back emf tries to oppose the change in current.
For a capacitor: -
$$I = Cdfracdvdt$$
This means that a current is produced when the voltage tries to change and that current tries to oppose the change in voltage.
And also want to know that when inductor suddenly open it generates
very huge voltage to keep current constant does it violate
conservation of energy ?
No, there is no violation of energy conservation. The amount of energy an inductor can use to produce a high back emf is limited by the current passing through the inductor when the terminals were open circuited. That energy is: -
$$W = dfrac12cdot LI^2$$
So, if the current was 10 amps when a 100 uH inductor open-circuited, the energy of the spark can be no more than 5 milli joules.
answered Aug 8 at 12:53
Andy aka
232k10172395
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
1
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
add a comment |Â
up vote
3
down vote
For an inductor: -
$$V = Ldfracdidt$$
This means a back emf voltage is produced when the current tries to change - that back emf tries to oppose the change in current.
For a capacitor: -
$$I = Cdfracdvdt$$
This means that a current is produced when the voltage tries to change and that current tries to oppose the change in voltage.
And also want to know that when inductor suddenly open it generates
very huge voltage to keep current constant does it violate
conservation of energy ?
No, there is no violation of energy conservation. The amount of energy an inductor can use to produce a high back emf is limited by the current passing through the inductor when the terminals were open circuited. That energy is: -
$$W = dfrac12cdot LI^2$$
So, if the current was 10 amps when a 100 uH inductor open-circuited, the energy of the spark can be no more than 5 milli joules.
answered Aug 8 at 12:53
Andy aka
232k10172395
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
1
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For an inductor: -
$$V = Ldfracdidt$$
This means a back emf voltage is produced when the current tries to change - that back emf tries to oppose the change in current.
For a capacitor: -
$$I = Cdfracdvdt$$
This means that a current is produced when the voltage tries to change and that current tries to oppose the change in voltage.
And also want to know that when inductor suddenly open it generates
very huge voltage to keep current constant does it violate
conservation of energy ?
No, there is no violation of energy conservation. The amount of energy an inductor can use to produce a high back emf is limited by the current passing through the inductor when the terminals were open circuited. That energy is: -
$$W = dfrac12cdot LI^2$$
So, if the current was 10 amps when a 100 uH inductor open-circuited, the energy of the spark can be no more than 5 milli joules.
answered Aug 8 at 12:53
Andy aka
232k10172395
For an inductor: -
$$V = Ldfracdidt$$
This means a back emf voltage is produced when the current tries to change - that back emf tries to oppose the change in current.
For a capacitor: -
$$I = Cdfracdvdt$$
This means that a current is produced when the voltage tries to change and that current tries to oppose the change in voltage.
And also want to know that when inductor suddenly open it generates
very huge voltage to keep current constant does it violate
conservation of energy ?
No, there is no violation of energy conservation. The amount of energy an inductor can use to produce a high back emf is limited by the current passing through the inductor when the terminals were open circuited. That energy is: -
$$W = dfrac12cdot LI^2$$
So, if the current was 10 amps when a 100 uH inductor open-circuited, the energy of the spark can be no more than 5 milli joules.
answered Aug 8 at 12:53
Andy aka
232k10172395
answered Aug 8 at 12:53
Andy aka
232k10172395
answered Aug 8 at 12:53
Andy aka
232k10172395
answered Aug 8 at 12:53
Andy aka
232k10172395
232k10172395
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
1
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
add a comment |Â
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
1
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
I know these equations already i want to know physical aspect
– user48391
Aug 8 at 17:11
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Thanks for answer of inductor not violating energy conservation that I don't know
– user48391
Aug 8 at 17:13
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
Explain me please what opposes voltage change across capacitor in detail sir
– user48391
Aug 8 at 17:14
1
1
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
At the borderline between the disciplines of electrical engineering and physics are the first two equations in my answer. To apply and use those equations is an electrical discipline, of which I count myself as an electrical engineer but, to get beneath those equations requires a deeper understanding of the physics and this is somewhat beyond the scope of this site. I use and work with the formulas and I recommend that you ask for a deeper understanding of these equations on the stack exchange physics site.
– Andy aka
Aug 8 at 18:17
add a comment |Â
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