Finding the step response of RL circuit

In summary, the conversation is about a circuit problem where the solution involves finding the steady state current provided by a source with a voltage of 18 V. The conversation also touches upon the use of u(t) in the solution and the manipulation of equations to find the current. There is a disagreement about the presence of a 6-ohm resistor in the circuit and its impact on the solution.
  • #1
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1

Homework Statement



A6mytYF.png


Homework Equations


The Attempt at a Solution



I understand everything except how they got i(∞). It seems like they used current division on 3A. But where are they applying current division and where did they get 3A? Also, in the last step, they magically added u(t). Where did they get that from?
 
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  • #2
Looks like they failed to specify that the magnitude of the step for Vs was 18 V.
 
  • #3
gneill said:
Looks like they failed to specify that the magnitude of the step for Vs was 18 V.

Oh, opps. That was my mistake. It looks like the problem is slightly different in two places.

Here's the problem:

qszXlHS.png


But I still don't see where they got the 3A and where u(t) in the last step came from.
 
  • #4
Supposing that the source voltage will be 18 V for "a long time", what current be the steady state current provided by the source?

The u(t) specifies that the solution pertains to time t > 0 since the source is stepped from 0 V to 18 V according to 18u(t) V.
 
  • #5
You can write (18-v)/6 = v(1/3 + 1/4).
So 3 = v(1/6 + 1/3 + 1/4) = v(1/2 + 1/4)

So there's your 3A on the left ...

and i = v/4 = 3/(4/2 + 4/4) = 3*2/(4+2) = 1A for i(infinity).

Just how he managed to get to 3*2/(4+2) immediately I don't know.
 
  • #6
gneill said:
Supposing that the source voltage will be 18 V for "a long time", what current be the steady state current provided by the source?

The u(t) specifies that the solution pertains to time t > 0 since the source is stepped from 0 V to 18 V according to 18u(t) V.

Would the below be correct?

09GIrQr.png


$$\frac { v-18 }{ 6 } +\frac { v-0 }{ 3 } +\frac { v-0 }{ 4 } =0\\ v=4\\ \\ i(∞)=\frac { v-0 }{ 4 } =1\quad A$$

rude man said:
You can write (18-v)/6 = v(1/3 + 1/4).
So 3 = v(1/6 + 1/3 + 1/4) = v(1/2 + 1/4)

So there's your 3A on the left ...

I can see how you manipulated that equation to get 3A on the left side but I don't understand what the right side means. On the right side, we have a sum of 3 currents: v/6, v/3, and v/4. v/3 is the current from v to ground through the 3-ohm resistor. v/4 is the current from v to ground through the 4-ohm resistor. But what is v/6? It can't be the current from v to ground through the 6-ohm resistor because a 6-ohm resistor doesn't exist between v and ground. I just don't see where the 3A current is on the circuit.
 
  • #7
Vishera said:
Would the below be correct?

09GIrQr.png


$$\frac { v-18 }{ 6 } +\frac { v-0 }{ 3 } +\frac { v-0 }{ 4 } =0\\ v=4\\ \\ i(∞)=\frac { v-0 }{ 4 } =1\quad A$$



I can see how you manipulated that equation to get 3A on the left side but I don't understand what the right side means. On the right side, we have a sum of 3 currents: v/6, v/3, and v/4. v/3 is the current from v to ground through the 3-ohm resistor. v/4 is the current from v to ground through the 4-ohm resistor. But what is v/6? It can't be the current from v to ground through the 6-ohm resistor because a 6-ohm resistor doesn't exist between v and ground. I just don't see where the 3A current is on the circuit.

You're looking for an actual 3A in the circuit where none exists. The 3A is just the result of rewriting my original equation which simply says that the current from the source to the v node = the sum of currents from the v node to ground.
 
  • #8
rude man said:
the current from the source to the v node = the sum of currents from the v node to ground.

I disagree here. On the right side, you had the term v/6 which is not a current from the v node to ground because there isn't a 6-ohm resistor between the v node and ground.
 
  • #9
Vishera said:
I disagree here. On the right side, you had the term v/6 which is not a current from the v node to ground because there isn't a 6-ohm resistor between the v node and ground.

"I do not agree with what you have to say, but I'll defend to the death your right to say it." :smile:
Voltaire
 

1. What is a step response of an RL circuit?

The step response of an RL circuit refers to the behavior of the circuit when a sudden change or "step" is applied to the input voltage. This can be visualized as a sudden increase or decrease in the input voltage and observing the resulting change in the output voltage.

2. How do you find the step response of an RL circuit?

To find the step response of an RL circuit, you need to first calculate the time constant of the circuit using the formula τ = L/R, where L is the inductance and R is the resistance. Then, you can use this time constant to plot the step response graph using the equation V(t) = V0(1-e^-t/τ), where V0 is the initial voltage and t is the time.

3. What is the significance of the step response in an RL circuit?

The step response in an RL circuit helps us understand the behavior of the circuit in response to sudden changes in the input voltage. It can also be used to analyze the stability and performance of the circuit, as well as to design and optimize circuit components.

4. How does the value of inductance affect the step response of an RL circuit?

The value of inductance plays a crucial role in determining the step response of an RL circuit. A higher inductance will result in a longer time constant, leading to a slower response to changes in the input voltage. On the other hand, a lower inductance will result in a shorter time constant and a faster response.

5. What are some applications of studying the step response of RL circuits?

The step response of RL circuits has many practical applications, such as in the design of electronic filters, power supplies, and motor control systems. It is also useful in understanding and troubleshooting the behavior of electrical systems in various applications, including telecommunications, robotics, and energy management.

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