Mixing
Impossible to find the energy stored in an inductor without resistance and given a constant DC current source?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Here's my circuit:
simulate this circuit – Schematic created using CircuitLab
After doing some series and parallel inductor combining I get the value $$L_eq=20.8 mH$$
My question is, how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
Here are some fundamental equations:
$$W_L(t)=1over2Li^2_L(t);;;;;;;V_L(t)=Ldi_L(t)over dt;;;;;;I_L(t)=1over Lint^t_t_0V_L(t)dt+I_L(t_0)$$
If I try to find the voltage using my equivalent inductance, then I will get zero since the derivative of a constant is zero, thus my energy anywhere will be zero.
I'm just curious to see if this is possible, thanks!
inductor parallel series
asked 5 hours ago
JustHeavy
1767
 |Â
show 3 more comments
up vote
1
down vote
favorite
Here's my circuit:
simulate this circuit – Schematic created using CircuitLab
After doing some series and parallel inductor combining I get the value $$L_eq=20.8 mH$$
My question is, how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
Here are some fundamental equations:
$$W_L(t)=1over2Li^2_L(t);;;;;;;V_L(t)=Ldi_L(t)over dt;;;;;;I_L(t)=1over Lint^t_t_0V_L(t)dt+I_L(t_0)$$
If I try to find the voltage using my equivalent inductance, then I will get zero since the derivative of a constant is zero, thus my energy anywhere will be zero.
I'm just curious to see if this is possible, thanks!
inductor parallel series
asked 5 hours ago
JustHeavy
1767
The problem is undefined as presented. With zero-resistance inductors, if one of them had a non-zero current initially, the total energy stored would be greater than zero. However, the total energy that you could get out of the thing would just be the traditional inductor energy equation using current and L_eq.
– TimWescott
5 hours ago
As far as inductors go, the equations I listed are as much as I know about them (and how to combine them). I'm not sure what reactance means here. Maybe saying no resistance was bad, I'm not explicitly saying that the circuit has no resistance, I just don't have any resistors in the circuit. My question could made into an example by asking what the energy stored in the 12mH inductor is.
– JustHeavy
5 hours ago
Do you know how inductors behave in DC circuits versus AC circuits?
– KingDuken
5 hours ago
No, I do not. That is what I am starting to think this circuit might apply to.
– JustHeavy
5 hours ago
1
Are you sure you got the schematic correct? As drawn, only L1 and L2 have any effect---the rest are shorted out. So the equivalent is 16 mH, not 20.8.
– The Photon
5 hours ago
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's my circuit:
simulate this circuit – Schematic created using CircuitLab
After doing some series and parallel inductor combining I get the value $$L_eq=20.8 mH$$
My question is, how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
Here are some fundamental equations:
$$W_L(t)=1over2Li^2_L(t);;;;;;;V_L(t)=Ldi_L(t)over dt;;;;;;I_L(t)=1over Lint^t_t_0V_L(t)dt+I_L(t_0)$$
If I try to find the voltage using my equivalent inductance, then I will get zero since the derivative of a constant is zero, thus my energy anywhere will be zero.
I'm just curious to see if this is possible, thanks!
inductor parallel series
asked 5 hours ago
JustHeavy
1767
Here's my circuit:
simulate this circuit – Schematic created using CircuitLab
After doing some series and parallel inductor combining I get the value $$L_eq=20.8 mH$$
My question is, how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
Here are some fundamental equations:
$$W_L(t)=1over2Li^2_L(t);;;;;;;V_L(t)=Ldi_L(t)over dt;;;;;;I_L(t)=1over Lint^t_t_0V_L(t)dt+I_L(t_0)$$
If I try to find the voltage using my equivalent inductance, then I will get zero since the derivative of a constant is zero, thus my energy anywhere will be zero.
I'm just curious to see if this is possible, thanks!
inductor parallel series
inductor parallel series
asked 5 hours ago
JustHeavy
1767
asked 5 hours ago
JustHeavy
1767
asked 5 hours ago
JustHeavy
1767
asked 5 hours ago
JustHeavy
1767
asked 5 hours ago
JustHeavy
1767
1767
The problem is undefined as presented. With zero-resistance inductors, if one of them had a non-zero current initially, the total energy stored would be greater than zero. However, the total energy that you could get out of the thing would just be the traditional inductor energy equation using current and L_eq.
– TimWescott
5 hours ago
As far as inductors go, the equations I listed are as much as I know about them (and how to combine them). I'm not sure what reactance means here. Maybe saying no resistance was bad, I'm not explicitly saying that the circuit has no resistance, I just don't have any resistors in the circuit. My question could made into an example by asking what the energy stored in the 12mH inductor is.
– JustHeavy
5 hours ago
Do you know how inductors behave in DC circuits versus AC circuits?
– KingDuken
5 hours ago
No, I do not. That is what I am starting to think this circuit might apply to.
– JustHeavy
5 hours ago
1
Are you sure you got the schematic correct? As drawn, only L1 and L2 have any effect---the rest are shorted out. So the equivalent is 16 mH, not 20.8.
– The Photon
5 hours ago
 |Â
show 3 more comments
The problem is undefined as presented. With zero-resistance inductors, if one of them had a non-zero current initially, the total energy stored would be greater than zero. However, the total energy that you could get out of the thing would just be the traditional inductor energy equation using current and L_eq.
– TimWescott
5 hours ago
As far as inductors go, the equations I listed are as much as I know about them (and how to combine them). I'm not sure what reactance means here. Maybe saying no resistance was bad, I'm not explicitly saying that the circuit has no resistance, I just don't have any resistors in the circuit. My question could made into an example by asking what the energy stored in the 12mH inductor is.
– JustHeavy
5 hours ago
Do you know how inductors behave in DC circuits versus AC circuits?
– KingDuken
5 hours ago
No, I do not. That is what I am starting to think this circuit might apply to.
– JustHeavy
5 hours ago
1
Are you sure you got the schematic correct? As drawn, only L1 and L2 have any effect---the rest are shorted out. So the equivalent is 16 mH, not 20.8.
– The Photon
5 hours ago
The problem is undefined as presented. With zero-resistance inductors, if one of them had a non-zero current initially, the total energy stored would be greater than zero. However, the total energy that you could get out of the thing would just be the traditional inductor energy equation using current and L_eq.
– TimWescott
5 hours ago
The problem is undefined as presented. With zero-resistance inductors, if one of them had a non-zero current initially, the total energy stored would be greater than zero. However, the total energy that you could get out of the thing would just be the traditional inductor energy equation using current and L_eq.
– TimWescott
5 hours ago
As far as inductors go, the equations I listed are as much as I know about them (and how to combine them). I'm not sure what reactance means here. Maybe saying no resistance was bad, I'm not explicitly saying that the circuit has no resistance, I just don't have any resistors in the circuit. My question could made into an example by asking what the energy stored in the 12mH inductor is.
– JustHeavy
5 hours ago
As far as inductors go, the equations I listed are as much as I know about them (and how to combine them). I'm not sure what reactance means here. Maybe saying no resistance was bad, I'm not explicitly saying that the circuit has no resistance, I just don't have any resistors in the circuit. My question could made into an example by asking what the energy stored in the 12mH inductor is.
– JustHeavy
5 hours ago
Do you know how inductors behave in DC circuits versus AC circuits?
– KingDuken
5 hours ago
Do you know how inductors behave in DC circuits versus AC circuits?
– KingDuken
5 hours ago
No, I do not. That is what I am starting to think this circuit might apply to.
– JustHeavy
5 hours ago
No, I do not. That is what I am starting to think this circuit might apply to.
– JustHeavy
5 hours ago
1
1
Are you sure you got the schematic correct? As drawn, only L1 and L2 have any effect---the rest are shorted out. So the equivalent is 16 mH, not 20.8.
– The Photon
5 hours ago
Are you sure you got the schematic correct? As drawn, only L1 and L2 have any effect---the rest are shorted out. So the equivalent is 16 mH, not 20.8.
– The Photon
5 hours ago
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
OK, so that's a trick question. Or a trick circuit, take your pick.
Let's start at the beginning. At some point the current through the circuit will be zero amps, and then the current source will be activated.
During turn-on, when the current source is connected or otherwise activated, the existence of a non-zero di/dt will produce non-zero voltages in all inductors which actually conduct current. Due to the short circuit across L3, there can be no voltage produced in L3 to L9, and consequently no current introduced into L3 to L9. With no instantaneous current at any time, the integral of the current in L3 to L9 must also be zero. So, at steady-state, only L1 and L2 have current flowing through them, and the current must be identically 10 amps in each.
From this you can calculate the total energy stored in all inductors.
answered 3 hours ago
WhatRoughBeast
48.1k22873
add a comment |Â
up vote
1
down vote
After doing some series and parallel inductor combining I get the value
Leq=20.8mH
You have miscalculated. All the inductors except L1 and L2 are shorted out, so the equivalent inductance is 16 mH.
how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
You can use the well-known formula
$$E = frac12LI^2$$
Since you have an ideal current source, there's no need to worry about the history of how the current reached this state; you know the current through each of the non-shorted inductors is 10 A.
So the energy stored in the 6 mH inductor is 0.3 J, and the energy stored in the 10 mH inductor is 0.5 J.
Now, in the absolute ideal of circuit theory, if these are perfect ideal inductors, it's possible for any amount of current to be circulating through the other (shorted out) inductors without affecting L1, L2, or the current source, and thus any amount of energy might be stored in them. You don't have any information to find out how much this energy is.
But really in circuit theory when we say we have an ideal inductor with "zero" resistance, we actually mean the resistance is too low to significantly affect any time constants or dynamic behavior in our circuit. But we know there is actually a non-zero resistance and so if this circuit is left to run for any appreciable amount of time, the small parasitic resistances of the inductors will absorb any energy in the shorted inductors, and they will have 0 current flowing.
On the other hand if this circuit was obtained by considering what happens when a switch is closed at some time $t_0$, you might be expected to use the configuration of the switch prior to the switch closing to determine the current through some of the shorted inductors.
answered 19 mins ago
The Photon
80.8k394191
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
OK, so that's a trick question. Or a trick circuit, take your pick.
Let's start at the beginning. At some point the current through the circuit will be zero amps, and then the current source will be activated.
During turn-on, when the current source is connected or otherwise activated, the existence of a non-zero di/dt will produce non-zero voltages in all inductors which actually conduct current. Due to the short circuit across L3, there can be no voltage produced in L3 to L9, and consequently no current introduced into L3 to L9. With no instantaneous current at any time, the integral of the current in L3 to L9 must also be zero. So, at steady-state, only L1 and L2 have current flowing through them, and the current must be identically 10 amps in each.
From this you can calculate the total energy stored in all inductors.
answered 3 hours ago
WhatRoughBeast
48.1k22873
add a comment |Â
up vote
4
down vote
OK, so that's a trick question. Or a trick circuit, take your pick.
Let's start at the beginning. At some point the current through the circuit will be zero amps, and then the current source will be activated.
During turn-on, when the current source is connected or otherwise activated, the existence of a non-zero di/dt will produce non-zero voltages in all inductors which actually conduct current. Due to the short circuit across L3, there can be no voltage produced in L3 to L9, and consequently no current introduced into L3 to L9. With no instantaneous current at any time, the integral of the current in L3 to L9 must also be zero. So, at steady-state, only L1 and L2 have current flowing through them, and the current must be identically 10 amps in each.
From this you can calculate the total energy stored in all inductors.
answered 3 hours ago
WhatRoughBeast
48.1k22873
add a comment |Â
up vote
4
down vote
up vote
4
down vote
OK, so that's a trick question. Or a trick circuit, take your pick.
Let's start at the beginning. At some point the current through the circuit will be zero amps, and then the current source will be activated.
During turn-on, when the current source is connected or otherwise activated, the existence of a non-zero di/dt will produce non-zero voltages in all inductors which actually conduct current. Due to the short circuit across L3, there can be no voltage produced in L3 to L9, and consequently no current introduced into L3 to L9. With no instantaneous current at any time, the integral of the current in L3 to L9 must also be zero. So, at steady-state, only L1 and L2 have current flowing through them, and the current must be identically 10 amps in each.
From this you can calculate the total energy stored in all inductors.
answered 3 hours ago
WhatRoughBeast
48.1k22873
OK, so that's a trick question. Or a trick circuit, take your pick.
Let's start at the beginning. At some point the current through the circuit will be zero amps, and then the current source will be activated.
During turn-on, when the current source is connected or otherwise activated, the existence of a non-zero di/dt will produce non-zero voltages in all inductors which actually conduct current. Due to the short circuit across L3, there can be no voltage produced in L3 to L9, and consequently no current introduced into L3 to L9. With no instantaneous current at any time, the integral of the current in L3 to L9 must also be zero. So, at steady-state, only L1 and L2 have current flowing through them, and the current must be identically 10 amps in each.
From this you can calculate the total energy stored in all inductors.
answered 3 hours ago
WhatRoughBeast
48.1k22873
answered 3 hours ago
WhatRoughBeast
48.1k22873
answered 3 hours ago
WhatRoughBeast
48.1k22873
answered 3 hours ago
WhatRoughBeast
48.1k22873
48.1k22873
add a comment |Â
add a comment |Â
up vote
1
down vote
After doing some series and parallel inductor combining I get the value
Leq=20.8mH
You have miscalculated. All the inductors except L1 and L2 are shorted out, so the equivalent inductance is 16 mH.
how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
You can use the well-known formula
$$E = frac12LI^2$$
Since you have an ideal current source, there's no need to worry about the history of how the current reached this state; you know the current through each of the non-shorted inductors is 10 A.
So the energy stored in the 6 mH inductor is 0.3 J, and the energy stored in the 10 mH inductor is 0.5 J.
Now, in the absolute ideal of circuit theory, if these are perfect ideal inductors, it's possible for any amount of current to be circulating through the other (shorted out) inductors without affecting L1, L2, or the current source, and thus any amount of energy might be stored in them. You don't have any information to find out how much this energy is.
But really in circuit theory when we say we have an ideal inductor with "zero" resistance, we actually mean the resistance is too low to significantly affect any time constants or dynamic behavior in our circuit. But we know there is actually a non-zero resistance and so if this circuit is left to run for any appreciable amount of time, the small parasitic resistances of the inductors will absorb any energy in the shorted inductors, and they will have 0 current flowing.
On the other hand if this circuit was obtained by considering what happens when a switch is closed at some time $t_0$, you might be expected to use the configuration of the switch prior to the switch closing to determine the current through some of the shorted inductors.
answered 19 mins ago
The Photon
80.8k394191
add a comment |Â
up vote
1
down vote
After doing some series and parallel inductor combining I get the value
Leq=20.8mH
You have miscalculated. All the inductors except L1 and L2 are shorted out, so the equivalent inductance is 16 mH.
how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
You can use the well-known formula
$$E = frac12LI^2$$
Since you have an ideal current source, there's no need to worry about the history of how the current reached this state; you know the current through each of the non-shorted inductors is 10 A.
So the energy stored in the 6 mH inductor is 0.3 J, and the energy stored in the 10 mH inductor is 0.5 J.
Now, in the absolute ideal of circuit theory, if these are perfect ideal inductors, it's possible for any amount of current to be circulating through the other (shorted out) inductors without affecting L1, L2, or the current source, and thus any amount of energy might be stored in them. You don't have any information to find out how much this energy is.
But really in circuit theory when we say we have an ideal inductor with "zero" resistance, we actually mean the resistance is too low to significantly affect any time constants or dynamic behavior in our circuit. But we know there is actually a non-zero resistance and so if this circuit is left to run for any appreciable amount of time, the small parasitic resistances of the inductors will absorb any energy in the shorted inductors, and they will have 0 current flowing.
On the other hand if this circuit was obtained by considering what happens when a switch is closed at some time $t_0$, you might be expected to use the configuration of the switch prior to the switch closing to determine the current through some of the shorted inductors.
answered 19 mins ago
The Photon
80.8k394191
add a comment |Â
up vote
1
down vote
up vote
1
down vote
After doing some series and parallel inductor combining I get the value
Leq=20.8mH
You have miscalculated. All the inductors except L1 and L2 are shorted out, so the equivalent inductance is 16 mH.
how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
You can use the well-known formula
$$E = frac12LI^2$$
Since you have an ideal current source, there's no need to worry about the history of how the current reached this state; you know the current through each of the non-shorted inductors is 10 A.
So the energy stored in the 6 mH inductor is 0.3 J, and the energy stored in the 10 mH inductor is 0.5 J.
Now, in the absolute ideal of circuit theory, if these are perfect ideal inductors, it's possible for any amount of current to be circulating through the other (shorted out) inductors without affecting L1, L2, or the current source, and thus any amount of energy might be stored in them. You don't have any information to find out how much this energy is.
But really in circuit theory when we say we have an ideal inductor with "zero" resistance, we actually mean the resistance is too low to significantly affect any time constants or dynamic behavior in our circuit. But we know there is actually a non-zero resistance and so if this circuit is left to run for any appreciable amount of time, the small parasitic resistances of the inductors will absorb any energy in the shorted inductors, and they will have 0 current flowing.
On the other hand if this circuit was obtained by considering what happens when a switch is closed at some time $t_0$, you might be expected to use the configuration of the switch prior to the switch closing to determine the current through some of the shorted inductors.
answered 19 mins ago
The Photon
80.8k394191
After doing some series and parallel inductor combining I get the value
Leq=20.8mH
You have miscalculated. All the inductors except L1 and L2 are shorted out, so the equivalent inductance is 16 mH.
how would you go about getting the energy stored in any one of the inductors, given the information in this post (no initial conditions or functions of t for the source current)
You can use the well-known formula
$$E = frac12LI^2$$
Since you have an ideal current source, there's no need to worry about the history of how the current reached this state; you know the current through each of the non-shorted inductors is 10 A.
So the energy stored in the 6 mH inductor is 0.3 J, and the energy stored in the 10 mH inductor is 0.5 J.
Now, in the absolute ideal of circuit theory, if these are perfect ideal inductors, it's possible for any amount of current to be circulating through the other (shorted out) inductors without affecting L1, L2, or the current source, and thus any amount of energy might be stored in them. You don't have any information to find out how much this energy is.
But really in circuit theory when we say we have an ideal inductor with "zero" resistance, we actually mean the resistance is too low to significantly affect any time constants or dynamic behavior in our circuit. But we know there is actually a non-zero resistance and so if this circuit is left to run for any appreciable amount of time, the small parasitic resistances of the inductors will absorb any energy in the shorted inductors, and they will have 0 current flowing.
On the other hand if this circuit was obtained by considering what happens when a switch is closed at some time $t_0$, you might be expected to use the configuration of the switch prior to the switch closing to determine the current through some of the shorted inductors.
answered 19 mins ago
The Photon
80.8k394191
answered 19 mins ago
The Photon
80.8k394191
answered 19 mins ago
The Photon
80.8k394191
answered 19 mins ago
The Photon
80.8k394191
80.8k394191
add a comment |Â
add a comment |Â
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%2felectronics.stackexchange.com%2fquestions%2f403288%2fimpossible-to-find-the-energy-stored-in-an-inductor-without-resistance-and-given%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
The problem is undefined as presented. With zero-resistance inductors, if one of them had a non-zero current initially, the total energy stored would be greater than zero. However, the total energy that you could get out of the thing would just be the traditional inductor energy equation using current and L_eq.
– TimWescott
5 hours ago
As far as inductors go, the equations I listed are as much as I know about them (and how to combine them). I'm not sure what reactance means here. Maybe saying no resistance was bad, I'm not explicitly saying that the circuit has no resistance, I just don't have any resistors in the circuit. My question could made into an example by asking what the energy stored in the 12mH inductor is.
– JustHeavy
5 hours ago
Do you know how inductors behave in DC circuits versus AC circuits?
– KingDuken
5 hours ago
No, I do not. That is what I am starting to think this circuit might apply to.
– JustHeavy
5 hours ago
1
Are you sure you got the schematic correct? As drawn, only L1 and L2 have any effect---the rest are shorted out. So the equivalent is 16 mH, not 20.8.
– The Photon
5 hours ago