Current through a single inductor

In summary, the conversation discusses a circuit with two inductors in parallel and a resistive network with a switch. After the switch is opened, the voltage and current for the inductors are found using the equations v(t)=96(e)^{-2t} and i(t)=12(e)^{-2t}, as well as the individual currents for each inductor using the equations i_1(t)= \frac{1}{5} \int_{0}^{t} 96(e)^{-2x}dx -8 and i_2(t)= \frac{1}{20} \int_{0}^{t} 96(e)^{-2x}dx -4. It is explained that the integral of
  • #1
modulation
2
0

Homework Statement



Ok, so this isn't a question of how to solve, but rather why the solution works.

Say you have a circuit with 2 inductors in parallel, a 5H and a 20H, in parallel to the inductors is a resistive network with a resistance of 8Ω, and a switch, which has been closed long enough for the inductors to reach DC steady state. The power source is not shown, but initially there is a current through each inductor, 8A through the 5H, and 4A through the 20H. When the switch is opened, the problem asks to find the voltage and current for the inductors, as well as the current through each individual inductor.


Homework Equations



So the solution for the voltage is [tex]v(t)=96(e)^{-2t}[/tex]
, and current for both inductors is [tex]i(t)=12(e)^{-2t}[/tex]

The solution to the current through the individual 5H inductor is
[tex]i_1(t)= \frac{1}{5} \int_{0}^{t} 96(e)^{-2x}dx -8[/tex]

and for the 20H inductor is
[tex]i_2(t)= \frac{1}{20} \int_{0}^{t} 96(e)^{-2x}dx -4[/tex]



The Attempt at a Solution



So what I do not understand is, how does the integral of the voltage, divided by the inductance, subtract the initial current, give the current through an individual inductor over time??
Really curious as to how this is possible mathematically and why??
Am I missing something?
 
Last edited:
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  • #2
That follows fundamentally from the V/I characteristics of an inductor.

[tex]\frac{di}{dt} = v/L[/tex]

so

[tex] i(t) = \frac{1}{L} \int_0^t v(\lambda) \, d\lambda + i(0)[/tex]

BTW. The sign of the voltage in your example is technically inconsistent with the direction you're using for current, hence the reason I used "+ i(0)" instead of the minus you had.
 
  • #3
So even though the voltage was derived from a combined inductance, you can simply divide by the individual inductance to get the current?
[tex]
\frac{di}{dt} = v/L
[/tex]

It is just odd , that first off the combined inductance is the product/sum, and then the voltage is the i(t)*R.
Strange that division can undo this. I have searched in my textbook for a more detailed proof of this but no luck.
 
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  • #4
modulation said:
So even though the voltage was derived from a combined inductance, you can simply divide by the individual inductance to get the current?
[tex]
\frac{di}{dt} = v/L
[/tex]

It is just odd , that first off the combined inductance is the product/sum, and then the voltage is the i(t)*R.
Strange that division can undo this. I have searched in my textbook for a more detailed proof of this but no luck.

Sometimes it's just easier to lump some components together before solving a circuit. This doesn't mean that the basic circuit laws don't still hold for each individual element however.

There are times when you do run into problems here, but this only when the voltages you need to analyze the individual elements get "swallowed up" in the lumping together process. This sometimes happens with Thevenin equivalents for example, but it's not an issue in this case.
 
  • #5




Thank you for your question. The solution provided is based on the mathematical relationship between voltage, current, and inductance. When an inductor is in a circuit, it resists changes in current, and this is represented by the inductance value. In this case, the inductors are in parallel, so the total inductance is the sum of the individual inductances (5H + 20H = 25H).

When the switch is opened, the inductors will try to maintain the same current as before, but since the circuit is now open, the current cannot flow through the resistive network. This results in a change in the magnetic field within the inductors, which induces a voltage in the circuit. This voltage is represented by the equation v(t)=96(e)^{-2t}.

Now, to find the current through each individual inductor, we use the equation i(t)=L(di/dt), where L is the inductance and di/dt is the rate of change of current. In this case, the rate of change of current is given by the integral of the voltage over time, divided by the inductance, as shown in the solution provided. This is due to the mathematical relationship between current and voltage in an inductor, where the current is directly proportional to the rate of change of voltage.

The initial current values (8A and 4A) are subtracted because they represent the current flowing through the inductors before the switch was opened. By subtracting them from the integral, we are essentially taking into account the initial current that was already present in the circuit.

I hope this helps clarify the mathematical reasoning behind the solution. Let me know if you have any further questions.
 

1. What is an inductor?

An inductor is a passive electronic component that stores energy in the form of a magnetic field. It is typically made up of a coil of wire and is used in circuits to resist changes in current flow.

2. How does current flow through a single inductor?

When a voltage is applied to an inductor, it creates a magnetic field around the coil. As the current flows through the inductor, the magnetic field also changes. This change in magnetic field induces a voltage in the opposite direction, which resists the change in current flow.

3. What is the relationship between current and voltage in an inductor?

The relationship between current and voltage in an inductor is governed by Faraday's law of induction. It states that the induced voltage is directly proportional to the rate of change of the magnetic field. Therefore, as the current changes, the induced voltage will also change in the opposite direction.

4. What factors affect the current through a single inductor?

The current through a single inductor is affected by the inductance of the coil, the resistance of the circuit, and the frequency of the current. A higher inductance value will result in a larger change in current for a given change in voltage, while higher resistance and frequency will result in a smaller change in current.

5. How is current through a single inductor represented in a circuit diagram?

The current through a single inductor is represented by a symbol that looks like a coiled wire. It is typically labeled with an "L" to indicate its inductance value. It is placed in series with the rest of the circuit, with the direction of current flow indicated by an arrow.

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