What are the implications of considering infinite first derivatives in circuits?

In summary: I in a coil?I'm not sure what your question is?In summary, the current increases steadily, the e.m.f opposing to that increase. The battery makes a work against emf of the coil; at the end, the current achieves it's final value.
  • #1
mcastillo356
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Hi
I've attached the image of the circuit. My first thought before connecting it: if I connect it, I must consider the fact that current cannot circulate suddenly, because a sudden change in the amount of current should imply an infinite value in [tex]\dfrac{\Delta{I}}{\Delta{t}}[/tex], and a infinite e.m.f. Instead, the current increases steadily, the e.m.f opposing to that increase. The battery makes a work against emf of the coil; at the end, the current achieves it's final value. The work required to establish a current [tex]I[/tex] in a coil is [tex]\dfrac{1}{2}LI^2[/tex].
Which are the maths beside [tex]\dfrac{\Delta{I}}{\Delta{t}}=\infty[/tex]?
 

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  • #2
I'm not sure what your question is?

The equations for this circuit are the Kirchoff Voltage Law, the voltage across the inductor and the resistor must add up to the battery voltage (after the switch is closed). These voltages can be expressed as a function of the current flowing around the circuit, which is the same for all components.

So: E = L⋅(di/dt) + i⋅R after the switch is closed. Normal initial conditions are i(0) = 0.

di/dt is essentailly the same as Δi/Δt when the Δ is very small.
 
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  • #3
Sorry, I think I've used the wrong notation, and the wrong question: Why does point out [tex]\dfrac{\Delta{I}}{\Delta{t}}\rightarrow{+\infty}[/tex] if nothing worked as you mention?
 
  • #4
Hum... I'm not sure why it says that.

The Inductor voltage Vl = L⋅(Δi/Δt). If Δi/Δt → ∞, then Vl → ∞. Theoretically this could happen if there was current flowing and you opened the switch. I guess their point is that the inductor current can't change instantly because that would require an infinite voltage, which is true.

BTW, mathematically this is the same thing as saying the voltage on a perfect capacitor can't change instantly because it would require infinite current.

This is the fundamental nature of these circuit elements.

edit: It's also similar to saying you can't stop a moving object instantly because that would require infinite force.
 
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  • #5
DaveE said:
Hum... I'm not sure why it says that.

Theoretically this could happen if there was current flowing and you opened the switch.
Not if you model the switch as some sort of capacitor with variable capacitance. Then the circuit essentially becomes an RLC circuit that passes from overdamped phase to underdamped and it will finally do exponentially decaying oscillations of the current. $$\frac{dI}{dt}$$ will not become infinite though it might become very large.
 
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  • #6
hmm,...I think this way
[tex]\displaystyle\lim_{\Delta{t}\rightarrow{0}}\Bigg(\dfrac{\Delta{I}}{\Delta{t}}\Bigg)=\displaystyle\lim_{\Delta{t}\rightarrow{0}}\Bigg(\dfrac{I_f-0}{\Delta{t}}\Bigg)=\infty[/tex]
 
  • #7
mcastillo356 said:
hmm,...I think this way
[tex]\displaystyle\lim_{\Delta{t}\rightarrow{0}}\Bigg(\dfrac{\Delta{I}}{\Delta{t}}\Bigg)=\displaystyle\lim_{\Delta{t}\rightarrow{0}}\Bigg(\dfrac{I_f-0}{\Delta{t}}\Bigg)=\infty[/tex]
I don't know what ##I_f## is. But your question has more to do with calculus than with electric circuits. How can ##\frac{dI}{dt}## be finite as ##\Delta{t}## approaches zero?

We don't know your background. Have you studied calculus?
 
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  • #8
Yes, I studied, but a long time ago. This year I've got to study it again if I want to pass the exam to access grade in maths by the spanish Uned (National University of at Distance Education). The topics are Maths, Physics, English, Spanish, and Literature. My poor background is to have read the books. The exams are in july, due to covid-19
 
  • #9
mcastillo356 said:
Summary:: I have a circuit with a power source, a resistance, and a coil, but still not connected: thoughts before connecting it

Which are the maths beside
 
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  • #10
Puf! It's too complicated. I think I will follow trying to pass my exams. Exercises proposed are my level. One day I may understand this thread. I'm sure of it. Thank you everybody!
 
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  • #11
Delta2 said:
Not if you model the switch as some sort of capacitor with very small capacitance. Then the circuit essentially becomes an underdamped RLC circuit and it will do exponentially decaying oscillations of the current. $$\frac{dI}{dt}$$ will not become infinite though it might become very large.
But then it's not a switch anymore. It's a switch with a capacitor.

You have to assume, in these theoretical, lumped element, circuit questions that people mean what they say in the schematics. Otherwise, you don't really know what the question is. This is especially true when you are teaching new material to students.
 
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  • #12
DaveE said:
But then it's not a switch anymore. It's a switch with a capacitor.
I think in real world a switch can be modeled by some sort of variable capacitor (that we are bringing its plates closer when we are closing the switch, or moving its plates further apart when we are opening the switch).
You have to assume, in these theoretical, lumped element, circuit questions that people mean what they say in the schematics. Otherwise, you don't really know what the question is. This is especially true when you are teaching new material to students.
However in theory a switch can be some sort of ideal device that instantly cuts the flow of current and creates an infinity on ##\frac{dI}{dt}##.
 
  • #13
Let's not head off into confusing territory for the OP please. No parasitic capacitance in this thread. If you want to discuss parasitic switch capacitance or explicit flyback HV transformer capacitances , please start a new thread. Thanks all.
 
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  • #14
berkeman said:
Let's not head off into confusing territory for the OP please. No parasitic capacitance in this thread. If you want to discuss parasitic switch capacitance or explicit flyback HV transformer capacitances , please start a new thread. Thanks all.
yes ok maybe i should not have bring in the modeling of a switch with a capacitor. In theory a switch is an ideal switch, no capacitance involved.
 
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  • #15
Delta2 said:
However in theory a switch can be some sort of ideal device that instantly cuts the flow of current and creates an infinity on ##\frac{dI}{dt}##.
Idealization in theory is the art to make things as simple as possible but not simpler (as Einstein put it). There's no way to cut the flow of current instantly, not even in the quasistatic approximations made for Kirchhoff circuit theory!
 
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  • #16
A simple rule IMO is that when a circuit schematic is shown, and Kirchhoff's laws are used, then the presumption is that all components are ideal.

Students taught circuit analysis can be introduced to the equivalent circuits of real life devices like switches and transformers, but the equivalent circuits are collections of ideal components and analyzed by ordinary CA.

So to introduce something like parasitic capacitance in a student-friendly way, redraw the circuit diagram.
 
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  • #17
Sure, but you have to use the realistic equivalent circuits and not something that leads to singularities.
 
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  • #18
vanhees71 said:
Sure, but you have to use the realistic equivalent circuits and not something that leads to singularities.
Now that's fun. I can't think of a circuit built from ideal passive components that leads to a singularity. Are you thinking of a short circuited ideal voltage source, or an open-circuited ideal current source?
 
  • #19
Yes, just describing a switch as a device which can produce a Heaviside unit-step function of the current is oversimplified and produces a singularity (Dirac ##\delta## distribution) for the time derivative. Another example is to "short-circuit" a charged capacitor without taking into account the finite resistance of the wire. Of course you can handle this with the theory of distributions somehow, but it's neither very physical nor simple for students to learn.
 
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  • #20
anorlunda said:
Now that's fun. I can't think of a circuit built from ideal passive components that leads to a singularity. Are you thinking of a short circuited ideal voltage source, or an open-circuited ideal current source?
Yes, I think you're correct. The singularities are usually from things you'd model with delta functions or steps. Like switches. It's an instantaneous change of state ( i.e. not matching boundary conditions between to otherwise linear solutions). Of course this includes changing the circuit topology.
 
  • #21
DaveE said:
Yes, I think you're correct. The singularities are usually from things you'd model with delta functions or steps. Like switches. It's an instantaneous change of state ( i.e. not matching boundary conditions between to otherwise linear solutions). Of course this includes changing the circuit topology.
It's partly semantic. I tell students that anything which leads to infinite V or I is absurd and should not be considered. But what about first derivatives? @vanhees71 points out that ideal step functions require a Dirac ##\delta## in the derivatives.

That's true, but considering ##V=L\frac{dI}{dt}## and ##I=C\frac{dV}{dt}## then singularities result in infinite V or I. Forbidding infinite V or I is sufficient, and separately forbidding singularities is redundant. I consider that a semantic difference.

But we don't want students to go overboard. Zero and infinite impedances are perfectly fine. We don't forbid all kinds of infinities in CA, just infinite V or I.
 
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  • #22
anorlunda said:
It's partly semantic. I tell students that anything which leads to infinite V or I is absurd and should not be considered. But what about first derivatives? @vanhees71 points out that ideal step functions require a Dirac ##\delta## in the derivatives.

That's true, but considering ##V=L\frac{dI}{dt}## and ##I=C\frac{dV}{dt}## then singularities result in infinite V or I. Forbidding infinite V or I is sufficient, and separately forbidding singularities is redundant. I consider that a semantic difference.

But we don't want students to go overboard. Zero and infinite impedances are perfectly fine. We don't forbid all kinds of infinities in CA, just infinite V or I.
Yes, it's all different ways of describing the same physics.
 

Related to What are the implications of considering infinite first derivatives in circuits?

1. How does an inductor store energy?

An inductor stores energy in the form of a magnetic field. When current flows through an inductor, the magnetic field around it increases. This magnetic field stores energy, which can be released when the current stops flowing.

2. What factors affect the amount of energy stored in an inductor?

The amount of energy stored in an inductor depends on its inductance, which is determined by the number of turns in the coil, the material of the core, and the size and shape of the coil. The current flowing through the inductor also affects the amount of energy stored.

3. Can energy be stored in an inductor indefinitely?

No, energy cannot be stored in an inductor indefinitely. The energy stored in an inductor is released when the current stops flowing. If the current continues to flow, the inductor will continue to store energy.

4. How is energy released from an inductor?

When the current through an inductor is interrupted, the magnetic field collapses and the energy stored in the inductor is released. This can happen when the power source is turned off or when the circuit is broken.

5. What are some practical applications of storing energy in an inductor?

Inductors are commonly used in electronic circuits to store energy and regulate current flow. They are also used in power grids to store energy and regulate voltage. Inductors are also used in devices like speakers and microphones to convert electrical energy into sound waves and vice versa.

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