Working Examples RLC Circuit with 5V Ramp Voltage Laplace Transforms

In summary, the conversation discussed the use of Laplace transforms to find the current function in series RLC circuits with a ramp voltage input. The solution involves solving for the current function using the conservation of voltage equation and then using fraction decomposition to put it in a form present in Laplace tables. The final answer is 7.5x10^-4[ 1 - e^-18.75t(cos 456.05t + 41.11x10^-3 sin 456.05t) ].
  • #1
Cooler
17
0
can anyone give me the working examples of current expressions of series RLC circuit using ramp voltage by Laplace transforms. thanks

R=15, L= 0.4H, C=12uF ...voltage 5v
 
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  • #2
You write a conservation of voltage around the loop:

[tex] -v(t) + Ri(t) + L\frac{di}{dt} + \frac{1}{c} \int i(t) \, dt=0[/tex]

Take the Laplace of both sides:

[tex] -V(s) + R I(s) + LsI(s) + \frac{1}{cs}I(s)= 0[/tex]

Solve for your sought after function of current:

[tex] I(s) = \frac{V(s)}{R + Ls + \frac{1}{cs}}[/tex]

Remember, this solution assumes initial conditions equal to zero.
 
  • #3
i stuck on the 1/s^2 [r + sL + 1/sC] ...do we need multiply all inside the bracket by s^2?
 
  • #4
Cooler said:
i stuck on the 1/s^2 [r + sL + 1/sC] ...do we need multiply all inside the bracket by s^2?

Yes. Then you will most likely need to do fraction decomposition. The trick will be putting the answer in a form present in your Laplace tables.
 
  • #5
i don't get the fraction decomposition...

when i multiplied inside the bracket it gives me...(15s^2 + 0.4s^3 + 83.333x10^3 s) is this correct and what to do next?
 
  • #6
Cooler said:
i don't get the fraction decomposition...

when i multiplied inside the bracket it gives me...(15s^2 + 0.4s^3 + 83.333x10^3 s) is this correct and what to do next?

[tex] I(s) = \frac{1}{s(Ls^2 + Rs + \frac{1}{c})}=-\frac{c^2(R+sL)}{cLs^2+cRs+1}+\frac{c}{s}[/tex]
 
  • #7
sorry i don't get the idea of how it end up that way...mind to explain? thnks
 
  • #8
Have you googled for tutorials yet? Google "fraction decomposition" or "partial fraction decomposition"

http://www.purplemath.com/modules/partfrac.htm

Essentially, it is a procedure for reversing the addition of two fractions.

so you can say a/b + c/d = (ad + bc)/(bd)

but in partial, you start with the right hand side and find the left hand side.Effectively (though it's not the complete picture), b = s and d = (ls^2 + rs + 1/c). And the reason I chose to only distribute 1 of the s in s^2 is that 1/s and 1/(as^2+bs+c) are usually in Laplace tables.
 
  • #9
What answer did you get?

my answer is 7.5x10^-4[ 1 - e^-18.75t(cos 456.05t + 41.11x10^-3 sin 456.05t)
 

Related to Working Examples RLC Circuit with 5V Ramp Voltage Laplace Transforms

1. What is an RLC circuit?

An RLC circuit is a type of electrical circuit that consists of a resistor (R), an inductor (L), and a capacitor (C). These components are connected in series or parallel and interact with each other to control the flow of electrical current.

2. 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 commonly used in electrical engineering to analyze and solve problems involving circuits, signals, and systems.

3. How does a 5V ramp voltage affect an RLC circuit?

A 5V ramp voltage is a type of input signal that increases linearly with time. When applied to an RLC circuit, it causes the current and voltage to change in a predictable manner, depending on the values of the R, L, and C components.

4. What are the advantages of using Laplace transforms in RLC circuit analysis?

Laplace transforms allow for the simplification of complex differential equations, making it easier to analyze the behavior of RLC circuits. They also provide a more intuitive understanding of the circuit's response to various input signals.

5. How can working examples of RLC circuits with 5V ramp voltage be useful in real-world applications?

RLC circuits are commonly used in electronic devices and systems, such as radios, televisions, and computers. By studying working examples with a 5V ramp voltage, scientists and engineers can better understand the behavior and performance of these circuits and apply that knowledge to design and improve real-world applications.

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