RC Circuit Potential and Current Calculation at Time=0 and Infinity

In summary, the conversation involved discussing a past quiz on a one-loop RC circuit problem. The circuit consisted of an ideal battery, a switch, two capacitors, and two resistors. The problem asked to calculate the potential and current at two points (A and B) at time=0 and at t=infinity. The attempt at a solution involved using Kirchoff's Loop Rule and a differential equation, but there were doubts about the accuracy of the solution due to a low score on the quiz. There was also a discussion about the application of Kirchoff's Loop Rule to RC circuits. It was suggested that the voltage of the capacitors should sum up to the voltage of the battery at t=infinity.
  • #1
Dahaka14
73
0

Homework Statement


Okay, I don't EXACTLY remember the problem statement or the circuit diagram, but I believe I can give enough information to do the problem. On this past quiz, I got 75/100, where there are 4 parts: 2 short answer, 1 part multiple choice, 1 part group test. I got 25 on MC, the group, and 24 on one of the short answer. This RC circuit question I got 1/25! I haven't gotten it back yet, but I thought I knew what I was doing.
There was a one-loop (obviously series) RC circuit, with, in order from first to last going counter-clockwise from the right side of the loop, an ideal battery, a switch, a capacitor, a resistor, another capacitor, then another resistor. Basically it went
(A)__R2___C2_(B)_R1__C1__ \__
l************************l
l************************+
l************************V
l************************-
l___________________________l
(Please ignored the asterisks, it was only a method applied to keep the spacing of the rectangle of the loop there in the post.)
Then it asked to calculate the potential and current at time=0 at (A) and (B), then at t=infinity.

Homework Equations


Kirchoff's Loop Rule: sum of voltages is 0
differential equation with charge term and time derivative of charge

The Attempt at a Solution


Current at t=0 is 0A for both (A) and (B) because, I forgot to put it down in the statement, but it said the capacitors are initially uncharged. I also said that the potential at both are equal to the battery's voltage, but now that I think about it, that doesn't sound plausible since there is no current. At t=infinity, I used Kirchoff's Loop Rule to solve for the current I
0=V-Q/C1-IR1-Q/C2-IR2 ---> I(t)=e^(-t/((R1+R2)C(eq)))
since the current should be everywhere the same on a series circuit. Then, for the potentials, I used Kirchoff's Loop Rule, but for (B) I took out the last two terms since the current "hasn't hit them" to drop the potential. I added the last two terms and solved for (B). Is this incorrect? I seem to be doubting myself now that I only got 1/25, which is like no partial credit. I filled the entire page with work. I also remember when calculating C(eq), I forgot to take 1/C(eq) after using 1/C(eq)=1/C1+1/C2. But that was used throughout the problem, so wouldn't that be double jeopardy to keep charging for that? I'm not sure, I need help please.
 
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  • #2
When a long time has passed, the capacitors are fully charged. How could current flow through C1 if it's fully charged?
 
  • #3
I may be wrong but I don't think its best to apply Kirchoff's Loop Rules all too heavily when it comes to RC circuitry. The thing about his method is that it applies to steady state circuits (not so much to dynamic circuits)...although the net voltage = 0V around the circuit seems right. The thing about RC circuits is that there is a current at first, but since capacitors are technically open ends in a circuit, charge simply accumulates on the capacitor plates producing an electric field across the capacitors and, as a result, a voltage. Once the voltage of the capacitors sums up to the voltage of the battery, at that point there is no more effective current along the circuit. The current at t = 0 at point A should be some value greater than 0 and at point B it should technically equal point A. The voltage from A to B could be given as a sum of the voltages across the capacitors and the voltage drop across the resistors; if the net voltage across a circuit is 0V, then the voltage between points A and B at t = 0 should be the voltage of the battery. As t approaches infinite, the voltage across the capacitors should sum up to the voltage across the battery, so it should technically remain constant.
 

1. What is an RC circuit and what is it used for?

An RC circuit, also known as a resistor-capacitor circuit, is a type of electrical circuit that contains a resistor and a capacitor. It is commonly used in electronic devices to control the flow of electric current and store electrical energy.

2. How does an RC circuit work?

An RC circuit works by using the resistor to limit the flow of current and the capacitor to store electrical energy. When the circuit is first connected, the capacitor charges up to the same voltage as the power supply. As the capacitor charges, the current decreases until it reaches zero. When the circuit is disconnected, the capacitor discharges and releases the stored energy.

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

The time constant of an RC circuit is a measure of how quickly the capacitor charges or discharges. It is calculated by multiplying the resistance value (in ohms) by the capacitance value (in farads). The time constant is used to determine the behavior of the circuit, such as how long it takes for the capacitor to reach a certain charge or discharge level.

4. What factors affect the time constant of an RC circuit?

The time constant of an RC circuit can be affected by the values of the resistor and capacitor, as well as the voltage of the power supply. A higher resistance or capacitance will result in a longer time constant, while a higher voltage will result in a shorter time constant. The arrangement of the components in the circuit can also affect the time constant.

5. How is an RC circuit tested?

To test an RC circuit, a voltage source and a multimeter are typically used. The circuit is connected to the voltage source and the multimeter is used to measure the voltage across the capacitor. The voltage should increase as the capacitor charges and decrease as it discharges. By measuring the time it takes for the voltage to change, the time constant of the circuit can be calculated and compared to the expected value.

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