# 3-Step Process of a Simple RC Circuit with Battery and an Open Switch

• jan2905
In summary, this conversation discusses a 3 step process involving a simple RC circuit with a battery and an initially open switch. The steps involve closing the switch to allow the capacitor to charge, inserting a dielectric material between the plates of the capacitor, and then opening the switch and removing the dielectric. The conversation also includes equations for calculating the change in potential energy of the capacitor during each step. There is some confusion about the notation used, but ultimately the change in potential energy in step 3 should be negative since the capacitor is able to discharge.
jan2905
The following 3 step process refers to a simple RC circuit with a battery and an initially open switch:

1- the switch is closed, allowing the capacitor to charge;
2- after the capacitor has charged, a slab of dielectric material is inserted between the plates of the capacitor and time passes;
3- the switch is opened, and the dielectric removed.

Q(capacitor)=CV(capacitor)
PE(capacitor)=1/2 CV^2

I said that: delta(PE)1>0; delta(PE)2>0; delta(PE)3<0

correct?

When you are putting the dielectric in the capacitor, you are effectively increasing the capacitance by the relationship that C/Co = ε the dielectric constant for the material, where ε > εo

As to your answers, I don't really know what the question is.

I would point out in step 3 that since the charge remains the same, but the capacitance is now influenced by εo instead of the higher ε of the removed dielectric, that the voltage will drop across the cap.

My question is whether or not the change in PE of the capacitor would reflect my "answer:" delta(PE)1>0; delta(PE)2>0; delta(PE)3<0 (the numerals are respective of the steps)

jan2905 said:
My question is whether or not the change in PE of the capacitor would reflect my "answer:" delta(PE)1>0; delta(PE)2>0; delta(PE)3<0 (the numerals are respective of the steps)

I guess I don't really understand your notation.

I just described what happens.

LowlyPion said:
I would point out in step 3 that since the charge remains the same, but the capacitance is now influenced by εo instead of the higher ε of the removed dielectric, that the voltage will drop across the cap.

You decrease ε so you increase V according to pion's equation and physical intuition - you would need a higher voltage to establish that charge without the dielectric!?

yeah, that is what my answer shows right? a drop in V means a drop in PE. so PE final - PE initial < 0 ?

epenguin said:
You decrease ε so you increase V according to pion's equation and physical intuition - you would need a higher voltage to establish that charge without the dielectric!?

That's correct. I misstated it in typing it.

Thanks for catching that.

soooooo, i am correct?

jan2905 said:
soooooo, i am correct?

No. I misstated my original description.

The equations were right. My English description was wrong.

ay... i don't understand what happens then. this is my logic:

1- capacitor is charging (PE final is greater than PE initial therefore deltaPE>0)
2- capacitor has dielectric inserted (PE final is greater than PE initial because dielectric allows more capacitance therefore deltaPE>0)
3- capacitor no longer charging and dielectric is removed (PE final is less that PE initial because capacitor is now able to discharge therefore deltaPE<0)

jan2905 said:
ay... i don't understand what happens then. this is my logic:

1- capacitor is charging (PE final is greater than PE initial therefore deltaPE>0)
2- capacitor has dielectric inserted (PE final is greater than PE initial because dielectric allows more capacitance therefore deltaPE>0)
3- capacitor no longer charging and dielectric is removed (PE final is less that PE initial because capacitor is now able to discharge therefore deltaPE<0)

3 does not correspond to 3 of your first post. Switch stays open, the charge has to stay there ideally. Yes, if the switch is then closed the capacitor will largely discharge.

yes 3 is the same on both posts:

3 (first post): the switch is opened, and the dielectric removed.
3 (last post): capacitor no longer charging [the switch is opened] and dielectric is removed

I am restating my logic. Is this correct? Anyone!

delta=final-initial; numerals correspond to the steps.

delta(PE)1>0; delta(PE)2>0; delta(PE)3<0

## 1. What is a simple RC circuit?

A simple RC (resistor-capacitor) circuit is a circuit that consists of a resistor and a capacitor connected in series. It is commonly used in electronic circuits to control the flow of electricity and store electrical energy.

## 2. What does the 3-step process of a simple RC circuit involve?

The 3-step process of a simple RC circuit involves charging, discharging, and steady state. Initially, when the switch is closed, the capacitor charges up to the same voltage as the battery. Then, when the switch is opened, the capacitor discharges through the resistor. Finally, the capacitor reaches a steady state where there is no more change in voltage.

## 3. How does a battery affect the behavior of a simple RC circuit?

The battery in a simple RC circuit acts as the power source, providing a constant voltage to charge the capacitor. Without the battery, the capacitor would not be able to charge and the circuit would not function.

## 4. What happens when the switch in a simple RC circuit is open?

When the switch is open in a simple RC circuit, the capacitor begins to discharge through the resistor. This creates a flow of current and causes the voltage to decrease until it reaches a steady state.

## 5. How does the value of the resistor and capacitor affect the behavior of a simple RC circuit?

The values of the resistor and capacitor in a simple RC circuit affect the time constant of the circuit, which is the time it takes for the capacitor to charge or discharge to 63.2% of its maximum voltage. A higher resistor value or a lower capacitor value will result in a longer time constant and vice versa.

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