Laplace transform of the given circuit

• Engineering
In summary: Equation 3:Vs = -I1 R2 + I2 (R2+R3)Vs = -10 + 20 I2 - 0,1 R2 + 0,1 R3 = 0Vs = 10 I1 - 0,1 R2 + 0,1 R3Vs = 10 I1 + 0,1 R2 - 0,1 R3In summary, the circuit has two Laplace domain equivalents and both equations have the same solution.
Homework Statement
Find i2(t) from given circuit
Relevant Equations
Laplace transform of R-L circuit
Hello i have an assignment. From given circuit i need to find s domain and inverse them back to t domain.
can you help me by explain this circuit?

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Show us your work. We are here to help with hints or to correct if you did it wrong. We can't do that unless we see your attempt at the solution first.

im sorry,but i really have no idea. i really don't know what to do.
please show me some hints first and i will try it and post it here

Our guidelines say this.
vela said:
Finally, we want you to develop the learning skills you need to succeed in all of your classes. You have to be able to get past the "I have no clue" phase on your own.

So if you were assigned this as homework, then the instructor thinks that you should be ready to solve it. What have you studied in this course about Laplace transforms? Did the lectures show how to analyze a circuit using Laplace properties?

JD_PM and berkeman
okay,understandable.
what do you think? where i miss?

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what do you think? where i miss?
Please learn to type your work into the forum window. That makes it *much* easier to read and for us to quote it. There is a tutorial at the top of the page under INFO, Help.

You can also see how LaTeX cleans up the posting of math equations in this similar thread:

As Berkeman suggested, try to type your work here.
If you insist on using pen and paper, re-draw the circuit, label the components and write the equations clearly, step by step.

Solution approach:
First, convert all the circuit components into their respective Laplace domain equivalents. Then you can go ahead with a circuit analysis method that you find convenient (node voltage method, mesh analysis, Thevenin/Norton etc).

berkeman
from mesh analysis,is this formula corect?

Equation 1:
I1 RE + (I1-I2) R2 - Vs = 0
I1 (R1+R2) - I2 R2 = Vs
4 = I1 (R1+R2) - I2 R2
4 = I1 (10+R2) - I2 R2
I1 = (4 + 10 I2)/40

Equation 2:
-I1 R2 + I2 (R2+R3) + Ldi2/dt = 0

put equation 1 to 2
-10 I1 + I2 (10+20) + 0,1 di
2/dt=0
-((4 + 10 I2)/40) + 30 I2 + 0,1 Si2(s) - S i (0)=0
-4 - 10 I2/40 + 30 I2 + 0,1 S i2(s) = 0
-0,1 - 0,25 I2 + 30 I2 + 0,1 I2
(s) = 0
-0,1 + 29,75 I2 + 0,1 I2 (s) = 0
- 0,1 + I2 (29,75 + 0,1 (s)) = 0
I2 = 0,1 / 29,75 + 0,1 (s)
I2 = 0,1/s / 29,75/s + 0,1 (s)

What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is often used in circuit analysis to simplify the analysis of circuits with time-varying signals.

How is a Laplace transform calculated?

A Laplace transform is calculated by integrating the function of time multiplied by the complex exponential function e^(-st), where s is a complex variable. The result is a function of s, which represents the complex frequency domain.

Why is a Laplace transform useful in circuit analysis?

A Laplace transform allows for the analysis of circuits with time-varying signals using algebraic equations instead of differential equations. This simplifies the analysis and makes it easier to solve for circuit variables such as voltage and current.

What is the inverse Laplace transform?

The inverse Laplace transform is the process of converting a function of s back into a function of time. It is the reverse process of the Laplace transform and is often used to find the time-domain solution of a circuit.

Are there any limitations to using Laplace transforms in circuit analysis?

Yes, there are some limitations to using Laplace transforms in circuit analysis. They are most effective for linear time-invariant circuits and may not be accurate for circuits with non-linear elements or time-varying components such as switches. Additionally, the use of Laplace transforms assumes the circuit is in a steady-state condition.

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