RC circuit and Laplace transforms

• darkfeffy
In summary: When you do the laplace transform, it will be solving for E(s) rather than 0, which is what you want.
darkfeffy
Hi everyone,

I know this should be obvious, but there's something I am just NOT getting.

Imagine a simple series RC circuit with a DC source as shown in the attachment. As can be seen from the picture, I have solved the differential equation in capacitor current in the time domain. In order to be able to solve the problem, I have assumed that dE/dt = 0 as this is a dc source.

I wish to know why I can't take the Laplace transform of both sides of the equation (*). I know that Laplace(0) = 0, so this would give a bogus equation (i.e. I(s) = 0, which is wrong). But if my equation (*) is right, then why can't I use the laplace transform of both sides at this point?

From textbooks, I read that the DC source is considered as a step input, thus in the Laplace domain, this would be E/s (in other words, the switch is closed at t=0). So, what is wrong with equation (*)?

Please don't hesitate to point out trivialities.

Thanks for your understanding.
e.

Attachments

• RC circuit.png
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$$\text{ Try }f(s) = \frac{1}{s-a} \space \text{ for } F(t) = 0 \text { for t <= 0 and } F(t)=e^{at} \text { for } t>0$$
There is no dc component for t>0. If you want a dc component, then try this online Laplace transform editor:

Put in 1 - ea*t and do Laplace to get
$$f(s) = \frac{1}{s}-\frac{1}{s-a}$$

First, if you are just using laplace transforms, you should be able to solve this with a lot less steps. The sooner you go into the laplace domain, the easier it is. You rewrite x(t) as X(s). If it is going to be applied as a step function, multiply your source by (1/s). Any function f(t), i(t), v(t) etc. becomes F(s), I(s) and if it is integrated, divide the function by s, and if it is differentiated, multiply by s. When you study the laplace transform in diff eq. it is treated as an operator, something you can multiply, that takes place of the integragtion and differentiation.

The answer to your question though, is that you are integrating zero on the right hand side, and when you integrate 0, you get a constant. To solve for this constant, simply set your time variable to 0, and you will see that your constant is equal to the current, which is equal to V/R. This gives you your voltage as the switch is closed. Substitue V/R into this constant and solve for your current now, and you will see your answer.

Last edited:
darkfeffy said:
Hi everyone,

In order to be able to solve the problem, I have assumed that dE/dt = 0 as this is a dc source.

If you assume dE/dt = 0 , doesn't that assume a steady state (i.e the situation long after the switch has been connected)?

Also, I thought I should add, when you go to the step of setting your dE/dt to zero, you are basically changing your circuit. How is this? Well, if you start from the very beginning with your first equation, and take the laplace transform, assuming a step input, and then differentiate by multiplying both sides by s in the laplace domain, you end up E(s), rather than 0.

By changing to dE/dt to 0, you are only considering your steady state, and basically removing the step function of the switch.

This is why I recommended starting the laplace transform as soon as possible without nulling out important parts of the equation like your dE/dt.

Last edited:
sophiecentaur said:
If you assume dE/dt = 0 , doesn't that assume a steady state (i.e the situation long after the switch has been connected)?

Yes, I think you're right that it is assuming steady state, and the differential equation * is setup as if the switch was already closed to begin with.

1. What is an RC circuit?

An RC circuit is a type of electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series. It is used for various purposes such as filtering, timing, and signal processing.

2. How does an RC circuit work?

An RC circuit works by charging and discharging the capacitor through the resistor. When a voltage is applied to the circuit, the capacitor charges up to the same voltage as the source. As the capacitor charges, the voltage across the resistor decreases until it reaches zero. This process repeats as long as the voltage is applied.

3. What is the time constant of an RC circuit?

The time constant of an RC circuit is the amount of time it takes for the capacitor to charge up to 63.2% of the applied voltage. It is calculated by multiplying the resistance (R) and the capacitance (C) of the circuit.

4. What is the role of Laplace transforms in RC circuits?

Laplace transforms are used in RC circuits to analyze the behavior of the circuit in the frequency domain. They help in solving differential equations that describe the circuit's behavior and provide a more convenient way to study the circuit's response to different inputs.

5. How do you calculate the transfer function of an RC circuit using Laplace transforms?

The transfer function of an RC circuit can be calculated by taking the Laplace transform of the circuit's differential equation and solving for the output voltage over the input voltage. This transfer function represents the relationship between the input and output signals of the circuit.

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