# DE explaining simple RC circuit

• bill.connelly
In summary, The person is trying to solve a simple system but is having trouble. They know there is no closed form solution and need to use convolution equations or a numerical approximation. They are trying to find the equation for dV/dt in terms of Vcmd, R1, R2, and C. They have two equations, Itot = V/R2 + C * dV/dt and V = Vcmd - Itot*R1, and rearrange them to get an equation for dV/dt. However, they are getting an issue with R1 approaching zero, and realize it is because of using an ideal voltage source and capacitor together. They are unsure if they are making an invalid assumption.
bill.connelly
I'm trying to wrap me head around what should be a very simple system, but I just can't manage it. I know there is no closed form for this problem (ignoring a voltage step like is shown in the diagram below, imagine it is a arbitrary voltage supply, or even more exactly, that R2 changes over time), and that to solve it I need to convolve equations (though I would hope I could do a pretty good numerical approximation too).

What I'm trying to do is get the equation for dV/dt in terms of Vcmd, R1, R2 and C

So I know
Itot = Ires + Icap
Itot = V/R2 + C * dV/dt

I also know V = Vcmd - Itot*R1
Itot = (Vcmd-V)/R1

Therefore

(Vcmd-V)/R1 = V/R2 + C * dV/dt
rearranging gives
C*dV/dt = (Vcmd-V)/R1 - V/R

I like that, because it means as R1 drops to zero, dV/dt approaches infinity (which is what I would expect, i.e. V should approach Vcmd instantly) BUT if R1 did drop to zero, and V = Vcmd, dV/dt should be zero, right? But that equation says otherwise. Likewise, as R1 approaches infinity, I would have thought dV/dt would approach zero.

Am I doing something wrong here? Making an invalid assumption?

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bill.connelly said:
I'm trying to wrap me head around what should be a very simple system, but I just can't manage it. I know there is no closed form for this problem (ignoring a voltage step like is shown in the diagram below, imagine it is a arbitrary voltage supply, or even more exactly, that R2 changes over time), and that to solve it I need to convolve equations (though I would hope I could do a pretty good numerical approximation too).

What I'm trying to do is get the equation for dV/dt in terms of Vcmd, R1, R2 and C

So I know
Itot = Ires + Icap
Itot = V/R2 + C * dV/dt

I also know V = Vcmd - Itot*R1
Itot = (Vcmd-V)/R1

Therefore

(Vcmd-V)/R1 = V/R2 + C * dV/dt
rearranging gives
C*dV/dt = (Vcmd-V)/R1 - V/R

I like that, because it means as R1 drops to zero, dV/dt approaches infinity (which is what I would expect, i.e. V should approach Vcmd instantly) BUT if R1 did drop to zero, and V = Vcmd, dV/dt should be zero, right? But that equation says otherwise. Likewise, as R1 approaches infinity, I would have thought dV/dt would approach zero.

Am I doing something wrong here? Making an invalid assumption?

Not sure why you think there is no closed form solution -- looks like your equations are just fine. The only issue is the infinity you think you are getting with R1 --> 0. You cannot use an ideal voltage source with zero output impedance and an ideal capacitor together in series with a voltage step. The non-real zero output impedance of the voltage source is what is giving you the non-real infinity.

I understand your confusion and frustration with trying to understand this simple RC circuit. Let me help clarify a few things for you.

First, it is important to note that the behavior of a circuit is dependent on the components used, such as the values of R1, R2, and C, as well as the input voltage, Vcmd. So, while there may not be a closed form solution for this problem, we can still use equations and numerical approximations to understand the behavior of the circuit.

Now, let's take a closer look at the equations you have written. You are correct in saying that Itot = V/R2 + C * dV/dt, as this is the sum of the current flowing through R2 and the current flowing through the capacitor. However, I believe there is a mistake in your second equation. It should be V = Vcmd - Itot * R1, as this takes into account the voltage drop across R1 due to the current flowing through it.

Now, let's rearrange the equations again:

Itot = (Vcmd - V)/R1
Itot = V/R2 + C * dV/dt

By substituting the first equation into the second, we get:

(Vcmd - V)/R1 = V/R2 + C * dV/dt
(Vcmd - V)*R2 = V*R1 + C * R1 * dV/dt

Now, we can use this equation to solve for dV/dt:

dV/dt = (Vcmd - V)*R2/(C * R1) - V/C

This equation makes more sense intuitively, as when R1 approaches infinity, the second term becomes negligible and dV/dt approaches (Vcmd - V)/R1, which is what we would expect. Similarly, as R1 approaches zero, the first term dominates and dV/dt approaches infinity.

I hope this helps clarify things for you. Remember, as a scientist, it is important to question and analyze your assumptions and equations to ensure they accurately represent the system you are studying. Keep up the good work!

## 1. What is a simple RC circuit?

A simple RC circuit consists of a resistor (R) and a capacitor (C) connected in series with a voltage source, typically a battery. The capacitor stores energy in the form of an electric field, while the resistor limits the flow of current through the circuit.

## 2. How does a simple RC circuit work?

In a simple RC circuit, the capacitor charges up when the circuit is closed, allowing current to flow through the circuit. As the capacitor charges, the voltage across it increases until it reaches the same voltage as the battery. Once the capacitor is fully charged, no more current flows through the circuit. When the circuit is opened, the capacitor discharges, releasing the stored energy back into the circuit.

## 3. What is the time constant of a simple RC circuit?

The time constant of a simple RC circuit is the product of the resistance (R) and the capacitance (C). It represents the amount of time it takes for the capacitor to charge or discharge to 63.2% of its maximum voltage or current.

## 4. How does the time constant affect the behavior of a simple RC circuit?

The time constant determines the rate at which the capacitor charges and discharges in the circuit. A larger time constant means the capacitor will take longer to charge and discharge, resulting in a slower response time for the circuit. A smaller time constant will result in a faster response time.

## 5. What are some real-life applications of simple RC circuits?

Simple RC circuits have a variety of practical applications, such as in filters, timing circuits, and signal conditioning. They are also commonly used in electronic devices, such as radios, televisions, and computers, to stabilize and regulate voltage levels.

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