RLC Circuit problem using laplace transforms

In summary, the switch at S is closed at time t=0 and a constant ... va−vh=v0 is applied. Kirchoff's laws were used to write 3 equations for i,i1,i2, and q(t). The relation between q(t) and i,i1,i2 was not determined. However, you can add the first two equations together to get a third equation.
  • #1
tkaz23
4
0

Homework Statement


The switch at S is closed at time t=0 and a constant ... va−vh=v0 is applied. Use Kirchoff's laws to write 3 equations for i,i1,i2, and q(t). What is the relation between q(t) and i,i1,i2? Given that the charge on C and all currents are initially zero, find an ODE for i1(t), given v0=1 volt, L=1 henry, C=1 farads, and R= 1 ohm.

Here is a picture of the RLC circuit (Hopefully this helps):
R i i_2
A___/\/\/\/\/___>__B___>___D
|aaaaaaaaaaaaaaaa | aaaaaaa|
aaaaaaaaaaaaaaa L $aaaaaaa C = q(t)
0 aaaaaaaaaaaaaaa $aaaaaaaa|
|_________________|_________|
H G F

Disclaimer: Disregard the blank a's. I am new to this, and I don't know how to get this to read white space. Sorry.

1st part asks to set equations for i,i_1,i_2, and q(t),
2nd part asks for a single equation for i_1,
3rd part asks for solution of part 2 using laplace transforms as well as expansions (solutions to the others)

Any help is appreciated, and thank you in advance. I have stared at this for quite sometime, and I have gotten no where.

Homework Equations


Kirchoff's 2 laws and Laplace Transform table


The Attempt at a Solution


This is what I got so far, but trying to use Laplace transforms on the i_1 eqn seems wrong because of the dq/dt in the left hand side.
I have i=i_1+i_2, i_2=dq=dt, and that the equations or differential equations are as follows for:
i_1) Ri_1+L(di_1\dt)=v_0(t)-R(dq/dt)
q) (1/C)q(t)+R(dq/dt)=v_0(t)-Ri_1
 
Last edited:
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  • #2
You need to use CODE tags to preserve spacing.

Code:
        R       i      i_2
A___/\/\/\/\/___>__B____>___D
|                  |        |
                 L $      C = q(t)
0                  $        |
|__________________|________|
H                  G        F
 
  • #3
Thank you for the pointing that out.
 
  • #4
tkaz23 said:

The Attempt at a Solution


This is what I got so far, but trying to use Laplace transforms on the i_1 eqn seems wrong because of the dq/dt in the left hand side.
I have i=i_1+i_2, i_2=dq=dt, and that the equations or differential equations are as follows for:
i_1) Ri_1+L(di_1\dt)=v_0(t)-R(dq/dt)
q) (1/C)q(t)+R(dq/dt)=v_0(t)-Ri_1
I take it i1 is the current through the inductor. If so, the equation ##i = i_1 + i_2## is fine.

I'm not sure how you got the other two equations. Start by first expressing the voltages VAB, VBG, and VDF in terms of i, i1, and q. If you go around the left loop (in a clockwise direction), KVL tells you that VAB+VBG+VGH+VHA = 0. What do you get when you substitute in the various values and expressions for those voltages?
 
  • #5
You get i+(di_1/dt)-1=0 for the left loop, q(t)-(di_1/dt)=0 for the right loop, and i+q(t)-1=0 for the entire loop, if you do what you say to do. Now, we pretty much have one equation. However when trying to use a laplace transform on this, I am unsure of what to do with i_2.
 
Last edited:
  • #6
I'm sorry to bump this, but I need to get someone to reply to me soon. If I hadn't stared at this thing for 4-5 hours, I would not be asking for help, and it is due in about an hour and a half. Please help! I am trying my best to understand things in a field, that I have relatively little clue of what to do.
 
  • #7
tkaz23 said:
You get i+(di_1/dt)-1=0 for the left loop, q(t)-(di_1/dt)=0 for the right loop, and i+q(t)-1=0 for the entire loop, if you do what you say to do. Now, we pretty much have one equation. However when trying to use a laplace transform on this, I am unsure of what to do with i_2.
Note you get your third equation if you add the first two equations together. It's not an independent equation, so you can toss it. Also, you should get in the habit of using variables and plugging the numbers in only at the end. So you have the equations
\begin{align*}
i_1 + i_2 &= i \\
Ri + L\frac{di_1}{dt} - V &= 0 \\
\frac{1}{C} q - L\frac{di_1}{dt} &= 0 \\
\end{align*}
To see how i2 works into this, you need to realize that i2 is the rate that charge flows into the capacitor. How do you express this mathematically?

tkaz23 said:
I'm sorry to bump this, but I need to get someone to reply to me soon. If I hadn't stared at this thing for 4-5 hours, I would not be asking for help, and it is due in about an hour and a half. Please help! I am trying my best to understand things in a field, that I have relatively little clue of what to do.
As you've discovered, it's best not to leave questions until the last minute. People offer help here on a volunteer basis, so while you'll often get a timely reply — that is, within a day or so — there's no guarantee that you'll get a quick reply.
 

FAQ: RLC Circuit problem using laplace transforms

1. What is an RLC Circuit?

An RLC circuit is an electrical circuit consisting of a resistor (R), an inductor (L), and a capacitor (C) connected in series. It is commonly used to study the behavior of electrical circuits and is often used in electronic devices.

2. How does Laplace transform help solve RLC circuit problems?

Laplace transform is a mathematical tool used to convert differential equations into algebraic equations, making it easier to solve complex problems. In RLC circuit problems, Laplace transform is used to analyze the behavior of the circuit over time and find the solutions for current and voltage.

3. What is the Laplace transform of a resistor, inductor, and capacitor?

The Laplace transform of a resistor is simply its resistance (R). The Laplace transform of an inductor is (Ls), where s is the complex frequency variable. The Laplace transform of a capacitor is (C/s), where s is the complex frequency variable.

4. How do I use Laplace transform to solve a second-order RLC circuit problem?

To solve a second-order RLC circuit problem using Laplace transform, you will need to first convert the circuit into a differential equation using Kirchhoff's laws. Then, apply Laplace transform to the differential equation and solve for the unknown variables. Finally, use the inverse Laplace transform to convert the solution back to the time domain.

5. What are the advantages of using Laplace transform in RLC circuit problems?

Laplace transform allows for the analysis of complex circuits and systems, making it a useful tool for solving RLC circuit problems. It also helps in reducing the complexity of the equations, making it easier to solve and understand the behavior of the circuit. Additionally, Laplace transform can be used to solve circuits with initial conditions, which is not possible with other methods.

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