Current in RL Circuit: Solving for Time-Dependent Current

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Discussion Overview

The discussion revolves around the time-dependent behavior of current in an RL circuit after a switch is closed, transitioning from steady state to a dynamic state. Participants explore the implications of the inductor's behavior during this transition and attempt to derive equations governing the current flow over time.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that at steady state, the inductor behaves like a short circuit, questioning what happens when the switch is closed and how the current changes over time.
  • Another participant clarifies that closing the switch disrupts the steady state, leading to a new dynamic condition.
  • There is a discussion about the initial current calculation at t=0, with one participant suggesting it equals V/R.
  • Some participants express confusion about the behavior of current as time approaches infinity, with one suggesting that current could approach infinity since it won't flow through the resistor.
  • Participants engage in deriving equations for current flow, with corrections and refinements made to earlier attempts, including the integration of voltage across the inductor.
  • There is a back-and-forth regarding the notation and meaning of variables in the equations, particularly the representation of current as i(t).
  • One participant acknowledges an earlier mistake in their equation and confirms that the correct expression for current builds up over time.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the circuit immediately after the switch is closed and the implications for current flow. There is no consensus on the behavior of current as time approaches infinity, with multiple interpretations presented.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the behavior of the inductor and the definitions of current at various time points. The discussion reflects varying levels of understanding and interpretation of circuit behavior.

Who May Find This Useful

Students and enthusiasts of electrical engineering or physics, particularly those studying RL circuits and transient analysis.

Tekneek
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Homework Statement



The circuit is in steady state before the switch is closed at t=0 I need to find how the current changes with time after the switch is closed.

The Attempt at a Solution



I read that at steady state inductor behaves like a short circuit, so does this mean when the switch closes all the current flows through the inductor? If so what happens after a long time has passed since the inductor opposes change in current flow? I am really confused...
 

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When the switch closes you no longer have steady-state. So no, the inductor does then no longer look like a short circuit.

But if the switch is left on long enough, you once again have a steady-state comndition.
 
Why wouldn't it be steady state?
 
Tekneek said:
Why wouldn't it be steady state?

Because you just flipped the switch, applying +10V to both components. Before that there was no voltage on either one.
 
rude man said:
Because you just flipped the switch, applying +10V to both components. Before that there was no voltage on either one.

So at t=0,

Current in the circuit = V/R = 10v/2kohms

and when t approaches infinity,

Current in the circuit = infinity? since the current won't flow through the resistor

Also, if i sum the current flow at the node (assuming all the currents are flowing out of the node)

-i(t) - V(t)/R - 1/L * Vdt from 0 to t

Does it look right?
 
Tekneek said:
So at t=0,

Current in the circuit = V/R = 10v/2kohms

and when t approaches infinity,

Current in the circuit = infinity? since the current won't flow through the resistor

Right!
Also, if i sum the current flow at the node (assuming all the currents are flowing out of the node)

-i(t) - V(t)/R - 1/L * ∫Vdt from 0 to t

Does it look right?

That's not an equation, and it's not the total outflow of current at the node either. (I sneaked an ∫ sign in for you).

Correct would be i(t) = V/R + (V/L)∫i dt from 0 to t.
 
rude man said:
Right!


That's not an equation, and it's not the total outflow of current at the node either. (I sneaked an ∫ sign in for you).

Correct would be i(t) = V/R + (V/L)∫i dt from 0 to t.


Thnx but where did u get I from?
 
Tekneek said:
Thnx but where did u get I from?

Same place you did. i(t) is the current shown in your diagram.

My equation says "current i(t) into the node = current out of the node."
 
rude man said:
Same place you did. i(t) is the current shown in your diagram.

My equation says "current i(t) into the node = current out of the node."

Sorry i meant the i in the integral part on the right hand side of the equation.
 
  • #10
Tekneek said:
Sorry i meant the i in the integral part on the right hand side of the equation.

i is shorthand for i(t).
 
  • #11
rude man said:
i is shorthand for i(t).

Isn't the equation for voltage across inductor,

V = L di/dt

Then when you integrate to find the current,

i = 1/L \intVdt = V/L \intdt

In your equation the current through the inductor is

i = V/L \inti dt, i don't get how there is an i there...
 
  • #12
Tekneek said:
Isn't the equation for voltage across inductor,

V = L di/dt

Then when you integrate to find the current,

i = 1/L \intVdt = V/L \intdt

In your equation the current through the inductor is

i = V/L \inti dt, i don't get how there is an i there...

The reason you don't get it is because I screwed up! :redface: You have the right expression.

So now you see how the current builds up to ∞ with time?
 
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  • #13
rude man said:
The reason you don't get it is because I screwed up! :redface: You have the right expression.

So now you see how the current builds up to ∞ with time?

lol. Well, the equation becomes

i(t) = v(t)/R + (v(t)/L)*t

When t is infinity i(t) is infinity, and when t=0, i(t)=V(t)/R

So plugging in everything i get,

i(t) = 0.005 + 1000t

Thanks for everything :)
 
  • #14
Congrats for getting the right answer despite my stumbling!
 

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