Understanding Charge Formulation in Discharging RC Circuits: A Quick Question

In summary, the original loop rule equation for charging a capacitor in a simple RC circuit (charged with one battery, one resistor and one capacitor all in series) has a negative sign that doesn't make sense. However, the books explanation is that the equation is the same as the one used for charging the capacitor just without the battery.
  • #1
Ja4Coltrane
225
0
So, I understand the charging of an RC circuit perfectly, but I am a tiny bit bothered by one part of the formulation of charge on a capacitor plate as a function of time for a discharging capacitor in a simple RC circuit (charged with one battery, one resistor and one capacitor all in series and discharged without the battery).
What bothers me is that the original loop rule equation seems to have a negative sign that doesn't make sense. The book gives the expression
-q/C-IR=0

I don't understand why both of these should be negative, in looking at the direction that the charge would flow, it seems that the equation should be

q/C-IR=0 or IR-q/C=0

The books explanation is that the formula is the same as the one used for charging the capacitor just without the battery, but this explanation doesn't quite make sense to me. I'm probably just being stupid. Is the reason perhaps that the -IR involves a negative value for I and therefore causes a positive potential difference?
 
Physics news on Phys.org
  • #2
Without a diagram it's hard to know for sure. But in writing loop equations, you generally follow the standard that a voltage increase (like going from the - to the + terminal of a voltage source as you trace the loop) is a positive voltage term, and a voltage drop (like when you trace in the direction of the current flow through a battery) is a negative voltage term in the equation. Does that help with your question?
 
Last edited:
  • #3
mmmm, not exactly. Here, imagine a simple circuit with no battery containing only a switch, a capacitor and a resistor in series. Consider that the capacitor is on the left of the resistor and the left plate is initialy (before the switch is closed) charged with a positive sign. The switch is then closed. It seems that the current should flow to the left (left for the resistor and capacitor). Because of this, if I do a loop rule it seems like the I would get +q/C-IR=0. Did that clear up my question?
 
  • #4
see here: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/RCSeries.html
 
Last edited by a moderator:

1. What is a Quick RC circuit?

A Quick RC circuit is a type of electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series. It is used to store and release electrical energy quickly.

2. How does a Quick RC circuit work?

When a voltage is applied to a Quick RC circuit, the capacitor charges up to the same voltage as the source. As the capacitor charges, the current through the circuit decreases until it reaches zero. When the capacitor is fully charged, it can quickly release the stored energy when the voltage is removed or when the circuit is completed.

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

The time constant of a Quick RC circuit is a measure of the rate at which the capacitor charges or discharges. It is calculated by multiplying the resistance (R) by the capacitance (C), represented by the equation τ = RC.

4. What is the practical application of a Quick RC circuit?

Quick RC circuits have various practical applications, including in electronic switches, filters, memory circuits, and time delay circuits. They are also commonly used in power supplies to filter out unwanted high-frequency signals.

5. How can I calculate the voltage across a capacitor in a Quick RC circuit?

The voltage across a capacitor in a Quick RC circuit can be calculated using the equation Vc = V0(1-e^(-t/RC)), where V0 is the initial voltage, t is the time, R is the resistance, and C is the capacitance. This equation can also be used to calculate the time it takes for the capacitor to charge or discharge to a certain voltage.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
789
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
204
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
6K
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top