Engineering Current in RL Circuit: Solving for Time-Dependent Current

[Tekneek]

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...

 

[rude man]

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.

 

[Tekneek]

Why wouldn't it be steady state?

 

[rude man]

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.

 

[Tekneek]

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?

 

[rude man]

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.

 

[Tekneek]

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?

 

[rude man]

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."

 

[Tekneek]

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.

 

[rude man]

Sorry i meant the i in the integral part on the right hand side of the equation.

i is shorthand for i(t).

 

[Tekneek]

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...

 

[rude man]

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! You have the right expression.

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

 

[Tekneek]

The reason you don't get it is because I screwed up! 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 :)

 

[rude man]

Congrats for getting the right answer despite my stumbling!