# Why would the inductor start conducting at t=0+ if impedance is infinte

#### Amnon Yaacobi

Joined Mar 10, 2022
14
Given a perfect theoretical un-charged inductor, to which we connect a voltage source across the terminals of the inductor, why would the inductor start conducting if at t=0+ the impedance of the inductor is infinite?
All sites and text books and forums just state that the inductor will start conducting without settling the contradiction that at t=0+ the impedance of the inductor is infinite.
I will be happy to learn

#### Papabravo

Joined Feb 24, 2006
21,225
How do you conclude that the impedance is infinite? The expression for impedance is:

$$Z_L\;=\;R_L+j\omega L$$

At zero frequency the DC resistance of the wire will pass a large but not infinite current. The time rate of change of the current times the inductance will determine the voltage across the inductor. Explain to me how this expression could ever be infinite. We live in the real world, not fantasyland.

Even if you assume that the DC resistance of the inductor is zero, there is resistance in the voltage source and the other circuit elements. An infinite current is simply not possible, neither is an infinite time rate of change of current. This means that even an ideal inductor cannot develop an infinite voltage across it or be subjected to an infinite current through it.

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#### Amnon Yaacobi

Joined Mar 10, 2022
14
as indicated there at t=0+ the an inductor behaves as an open circuit and they go about to explain why. So if that is the case, why would a an uncharged theoretical perfect inductor, connected to a theoretical perfect voltage source start conducting? may be it will not? I do not know and would love to hear an learned answer

#### Papabravo

Joined Feb 24, 2006
21,225
as indicated there at t=0+ the an inductor behaves as an open circuit and they go about to explain why. So if that is the case, why would a an uncharged theoretical perfect inductor, connected to a theoretical perfect voltage source start conducting? may be it will not? I do not know and would love to hear an learned answer
My understanding of inductors, real and otherwise, has stood the test of time, and I'm not strongly motivated to look into the matter. How does the author go on to explain what happens next in the case of an infinite impedance? What breaks the logjam?
Ultimately an inductor is governed by a differential equation and infinities are definitely not part of the process. They contribute very little to the understanding and are confusing to the uninitiated as well.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
My understanding of inductors, real and otherwise, has stood the test of time, and I'm not strongly motivated to look into the matter. How does the author go on to explain what happens next in the case of an infinite impedance? What breaks the logjam?
By no means do I question your understanding. I'll try to sum up what is written in that site (which I have seen in other references as well). If the inductor resists a change in current then at t=0+ we will have a voltage with no current therefore V/I=V/0=infinity.
And so my question still remains, why would the inductor start conducting?

#### Alec_t

Joined Sep 17, 2013
14,313
If the inductor resists a change in current
Resisting it is not the same as completely preventing it.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
Resisting it is not the same as completely preventing it.
But wouldn't infinite impedance will?

#### Papabravo

Joined Feb 24, 2006
21,225
By no means do I question your understanding. I'll try to sum up what is written in that site (which I have seen in other references as well). If the inductor resists a change in current then at t=0+ we will have a voltage with no current therefore V/I=V/0=infinity.
And so my question still remains, why would the inductor start conducting?
This is why they cause a problem. You are hung up on a meaningless proposition.

#### Papabravo

Joined Feb 24, 2006
21,225
But wouldn't infinite impedance will?
If you can find one, then I guess we can test the proposition. Until then I remain skeptical that resolving the issue has any practical importance.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
If you can find one, then I guess we can test the proposition. Until then I remain skeptical that resolving the issue has any practical importance.
Of course it has no practical importance. It does go toward deepening our understanding of inductors. As I stressed out in my original question, I am considering a theoretical inductor. As you indicated, a real inductor will have a finite R component and therefore will definitely conduct. I am very much still interested in an answer. Thank you for responding though. I cherish a good discussion

#### BobTPH

Joined Jun 5, 2013
8,958
Throw a ball in the air. Eventually it reaches its peak and the velocity is zero. Call that time t = 0. Does this mean the ball will not fall?

The answer is that, in both cases, the first derivative is non-zero. So immediately after t = 0 the velocity or current is non zero.

#### Alec_t

Joined Sep 17, 2013
14,313
You are hung up on a meaningless proposition.
Agreed.
Taking the proposition a bit further:
If i=0, then di/dt must be 0 at t=0 if no current can ever flow.
For any non-zero inductance L, E = L*di/dt = 0 at t=0.
Hence there's no back-emf E to resist current flow.
So we have a hypothetical inductor with an infinite impedance which doesn't resist current flow!
It's rather like asking what is 0/0, or what is 0 x ∞.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
Throw a ball in the air. Eventually it reaches its peak and the velocity is zero. Call that time t = 0. Does this mean the ball will not fall?

The answer is that, in both cases, the first derivative is non-zero. So immediately after t = 0 the velocity or current is non zero.
Is what written here https://testbook.com/question-answe...-initial-conditions--61bc449e3b40798b1aded48f wrong?
I think not. The ball has no impedance holding it back. The current through the (perfect) inductor at t=0+ does.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
Agreed.
Taking the proposition a bit further:
If i=0, then di/dt must be 0 at t=0 if no current can ever flow.
For any non-zero inductance L, E = L*di/dt = 0 at t=0.
Hence there's no back-emf E to resist current flow.
So we have a hypothetical inductor with an infinite impedance which doesn't resist current flow!
It's rather like asking what is 0/0, or what is 0 x ∞.
O.K. Now we are getting somewhere. Thank you for responding. So from 1st principals, if we apply a non-zero voltage E, across the inductor, it may be considered as applying a voltage step function or a change in voltage which is dE/dt. that entails that there must be a second derivative to the current, d2i/dt2 which is non zero. What does that tell us?

#### BobTPH

Joined Jun 5, 2013
8,958
The concept of impedance of a capacitor or inductive applies only in AC response to a sine wave.

Impedance at DC only applies to resistors. You (he) is mis-applying a concept that does not apply.

It really is exactly the same as a force and a mass. A mass resists velocity in the same was as an inductor resists current. The equation identical:

F = m dV / dt

V = L dI / dt

If we apply a force to a non-moving object, we do not say that it has infinite resistance to motion because it has zero velocity with non zero force.

Here is another way to see that the interpretation is wrong. He is saying that the impedance is voltage / current, right?

What if we have an inductor that has 1A flowing at t = 0. And we apply a voltage of 10V. Yes, you can do that, just like you can apply a force to a moving object. Is the impedance of that same inductor now 10 Ohms? And if we apply a voltage of 100V its impedance 100 Ohms? Clearly, the impedance is meaningless, because we can make it anything we want.

Furthermore, mathematically, 10 / 0 is undefined, not infinite. It is undefined because it leads to absurdities as above.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
Thank you all for the discussion
The concept of impedance of a capacitor or inductive applies only in AC response to a sine wave.

Impedance at DC only applies to resistors. You (he) is mis-applying a concept that does not apply.

It really is exactly the same as a force and a mass. A mass resists velocity in the same was as an inductor resists current. The equation identical:

F = m dV / dt

V = L dI / dt

If we apply a force to a non-moving object, we do not say that it has infinite resistance to motion because it has zero velocity with non zero force.

Here is another way to see that the interpretation is wrong. He is saying that the impedance is voltage / current, right?

What if we have an inductor that has 1A flowing at t = 0. And we apply a voltage of 10V. Yes, you can do that, just like you can apply a force to a moving object. Is the impedance of that same inductor now 10 Ohms? And if we apply a voltage of 100V its impedance 100 Ohms? Clearly, the impedance is meaningless, because we can make it anything we want.

Furthermore, mathematically, 10 / 0 is undefined, not infinite. It is undefined because it leads to absurdities as above.
Thank you. I think things are starting to clear up.
Building on what you wrote, and what i think is correct regarding dE/dt I would like to think about it somemore

#### Papabravo

Joined Feb 24, 2006
21,225
It might be handy to stop using the notion that something can "equal" infinity. The more correct way to say it is to use the language of limits. That is, the limit of 1/x as x approaches 0+ is infinite.

#### Amnon Yaacobi

Joined Mar 10, 2022
14
It might be handy to stop using the notion that something can "equal" infinity. The more correct way to say it is to use the language of limits. That is, the limit of 1/x as x approaches 0+ is infinite.
Thank you, I like that.
And thanks to all the other responders.
As I said, my goal was to deepen my understanding of inductors. I'll ponder some on this and post my thoughts soon

#### WBahn

Joined Mar 31, 2012
30,055
as indicated there at t=0+ the an inductor behaves as an open circuit and they go about to explain why. So if that is the case, why would a an uncharged theoretical perfect inductor, connected to a theoretical perfect voltage source start conducting? may be it will not? I do not know and would love to hear an learned answer
That site is simply trying to make a simple, superficial explanation without using things like calculus in the discussion based on the assumption that the math background of the target audience won't let them deal with those concepts.

The two main concepts that everything has to conform to is that, for an inductor, the constitutive equation that governs the relationship between voltage and current is:

v(t) = L · di(t)/dt

and that energy stored in the magnetic field of an inductor is

u(t) = L · i²(t) / 2

Because energy must be a continuous function of time, i(t) cannot change instantaneously, thus

i(t-) = i(t+)

It's important to keep in mind that this statement is talking about limits and is a shorthand way of saying that the limit of the current as we approach any particular time t from the left is equal to the current as we approach that same time from the right. This is satisfied as long as i(t) is a continuous function of time, it does NOT imply that i(t) can't change, or even change quite rapidly, only that it can't change instantaneously.

The claim that the inductor behaves like an open circuit at t=0+ is only true if the inductor had zero current through it at t=0-.

Dividing the instantaneous voltage across an inductor by the instantaneous current through it yields a meaningless quantity -- an inductor is NOT a resistor and it does NOT obey Ohm's Law. Impedance is a frequency-domain concept, not a time-domain one.

#### Ya’akov

Joined Jan 27, 2019
9,150
In any case where we encounter a discontinuity—infinity or zero—in a case where practical experience tells us it can’t be correct, the only deepening of understanding we can get is one of the limits of the mathematical description which is clearly not accounting for something the physical world includes.

The component with infinite or zero value of any parameter doesn’t exist. All practical components will include parasitics that are more than inconveniences for mathematicians. They are part of the fundamental nature of how things actually work.

The phantom ideal component is as useful as any theoretical proposition of perfection—it simplifies the math. It makes possible generalizations that are very useful in practical applications. But always keep in mind it is the fabricated “ideal” component that is the limited version of things, not the practical, actual component that can never be the ideal version.

Pursuing an extrapolation of the ideal as a way of learning more about the practical is upside down. You won’t learn anything useful from the simplifed model when you have reached its floor of precision.