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

• Engineering
• Tekneek
In summary: Well, the equation becomes i(t) = V/L \inti dt, which is correct.Thanks for everything :)Congrats for getting the right answer despite my stumbling!
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...

#### Attachments

• inductor.gif
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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.

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).

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 $\int$Vdt = V/L $\int$dt

In your equation the current through the inductor is

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

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 $\int$Vdt = V/L $\int$dt

In your equation the current through the inductor is

i = V/L $\int$i 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?

1 person
rude man said:
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 :)

Congrats for getting the right answer despite my stumbling!

## 1. What is an RL circuit?

An RL circuit is a type of electrical circuit that contains a resistor (R) and an inductor (L) connected in series. The inductor stores energy in the form of a magnetic field, while the resistor dissipates energy in the form of heat.

## 2. What is the equation for current in an RL circuit?

The equation for current in an RL circuit is I(t) = I0e-t/τ, where I0 is the initial current, t is time, and τ is the time constant, which is equal to L/R.

## 3. How do you solve for time-dependent current in an RL circuit?

To solve for time-dependent current in an RL circuit, you first need to determine the time constant (τ) using the formula L/R. Then, plug this value into the equation I(t) = I0e-t/τ along with the initial current (I0) and solve for the current at any given time (t).

## 4. What is the significance of the time constant in an RL circuit?

The time constant (τ) in an RL circuit represents the rate at which the current decreases in the circuit. It is a measure of the inductor's ability to store energy and the resistor's ability to dissipate energy. A larger time constant indicates a slower decrease in current, while a smaller time constant indicates a faster decrease.

## 5. Can the current in an RL circuit ever reach 0?

In theory, the current in an RL circuit will never reach 0, as it will continue to decrease exponentially. However, in practical applications, the current will eventually become small enough to be considered negligible for most purposes. This is typically referred to as the "steady-state" condition.

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