Differential equation in simple RC-circuit

In summary, the conversation is about a person trying to solve a problem in a series RC circuit with a DC source. They have solved the differential equation in capacitor current in the time domain and are wondering why they cannot take the Laplace transform of both sides of the equation. They also ask about the Laplace transform of a DC source and why their equation in the Laplace domain is incorrect. The response suggests that the Laplace transform of the equation should be sI + 1/RC I = 0, and the person should obtain the same answer by solving for I and inverting it.
  • #1
darkfeffy
17
0
Hi guys,

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.

What I wish to know is why can't I take the Laplace transform of both sides of the equation (*)? I know that L(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. So again, what is wrong with equation (*)? And why do I get the right answer at the end?

Again, as I said, I think this should be obvious, so please don't hesitate to point out trivialities.

Thanks for your understanding.
e.
 

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  • #2
Laplace transform of your equation (*) should be
[tex]sI+\frac{1}{RC}I=0[/tex]

Solve for I and invert it. You should obtain the same answer.
 

1. What is a differential equation in a simple RC-circuit?

A differential equation in a simple RC-circuit is a mathematical equation that describes the relationship between the voltage and current in a circuit containing a resistor (R) and a capacitor (C). It takes into account the rate of change of voltage and current over time, as well as the properties of the components in the circuit.

2. How is a differential equation derived for an RC-circuit?

The differential equation for an RC-circuit is derived using Kirchhoff's voltage law, which states that the sum of voltages around a closed circuit is equal to zero. By applying this law to an RC-circuit, we can derive a differential equation that describes the behavior of the circuit.

3. What are the applications of differential equations in RC-circuits?

Differential equations in RC-circuits have various applications in electrical engineering and physics. They can be used to analyze the behavior of circuits, design filters, and model the charging and discharging of capacitors in electronic systems.

4. How do initial conditions affect the solution of a differential equation in an RC-circuit?

Initial conditions, such as the initial voltage and current values, can affect the solution of a differential equation in an RC-circuit. These conditions determine the starting point of the solution and can impact the behavior of the circuit over time.

5. Are there any simplifications that can be made when solving a differential equation in an RC-circuit?

Yes, there are simplifications that can be made when solving a differential equation in an RC-circuit. For example, if the circuit is in steady state, the time derivative of the current and voltage can be assumed to be zero, simplifying the equation. Additionally, approximations can be made for certain components, such as assuming an ideal capacitor with no resistance or leakage.

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