RC time Circuit; charge of capacitor as a function of time derivation

In summary: Draw the circuit. What will be the potential drop across the capacitor and resistor respectively? What is the potential gain across the battery? How are those related?If we take the gain of the battery and subtract the potential drop across the resistor, we get the potential gain across the capacitor. This is the same as the potential drop across the capacitor because we are dividing by the gain of the battery.
  • #1
sobie925
2
0
Hey guys! New to physicsforums. I wanted to ask a more conceptual question regarding RC time Circuits. I spent some time trying to derive the equations, and I feel like I'm not setting up the problem correctly. Here's my attempt:

Solutions according to profecssor:

1) ##q_{charge}(t)=C\epsilon(1-e^{-\frac{t}{\tau}})##
2) ##q_{discharge}(t)=q_{0} e^{\frac{-t}{\tau}}##

---
my attempt at solution 1
---

##q##-charge on capacitor
##R##-Resistance encountered in circuit
##C##-Capacitance of capaticor
##\epsilon##-Electromotive Force of battery
##t## -time since capacitor began charging
##\tau## - ##RC##
##i## -current


##C=\frac{q}{\epsilon}##
∴##q=C\epsilon##
##\epsilon=iR=\frac{dq}{dt}R##
∴##q=RC\frac{dq}{dt}=\tau\frac{dq}{dt}##
∴##\frac{1}{q}\frac{dq}{dt}=\frac{1}{\tau}##
∴##\int_0^t \frac{1}{q}\frac{dq}{dt} dt=\int_0^t \frac{1}{\tau} dt##
∴##\int_{q(0)}^{q(t)} \frac{dq}{q}=\int_0^t \frac{1}{\tau} dt##
∴##ln(q(t))-ln(q(0))=\frac{t}{\tau}##
∴##ln(\frac{q(t)}{q(0)})=\frac{t}{\tau}##
∴##e^{\frac{t}{\tau}}=\frac{q(t)}{q(0)}##
∴##q(t)=q(0)e^{\frac{t}{\tau}}##
So it's here that I'm like "I guess ##q(0)=c\epsilon## just for the sake of looking more like the professor's solution" <I realize how stupid this is... lol
∴##q(t)=C\epsilon e^{\frac{t}{\tau}}##

There must be something that I'm just totally missing here. The math seems right, so I'm thinking it's the set-up that I'm missing.

Any help is appreciated! Thank you!
 
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  • #2
You have assumed that the potential drop across the resistance is the same as that across the capacitor. Is this true or can you think of something else that is more physical?
 
  • #3
Alright, so I started off on the wrong part. So, I need to derive ##V_{C}(t)=\epsilon(1-e^{\frac{-t}{\tau}})## (This was written on the paper our professor gave us, so as of now it doesn't really mean anything to me.)

So, ##V_{C}=iR##. The current is the same as the current we've been dealing with, but what R is this equation referring to?
 
  • #4
Let us start studying the charging circuit: Draw the circuit. What will be the potential drop across the capacitor and resistor respectively? What is the potential gain across the battery? How are those related?

For the discharging circuit: it is the same just without the battery.
 
  • #5



Hi there! Welcome to physicsforums. Your attempt at deriving the equation for the charge on a capacitor as a function of time is on the right track. However, there are a few things that could be clarified or corrected.

Firstly, your use of the equation ##C=\frac{q}{\epsilon}## is not necessary in this case. This equation is typically used to calculate the capacitance of a capacitor, given the charge and voltage. In this problem, the capacitance is given and we are trying to find the charge as a function of time.

Secondly, you do not need to assume that ##q(0)=C\epsilon##. This is because we are considering the charge on the capacitor as it is charging, and at the start, the charge is 0. So, we can simply let ##q(0)=0##.

Lastly, when integrating both sides of the equation, you should use the correct limits. In this case, the limits should be from 0 to t, as we are considering the charge at time t.

With these corrections, your derivation will become:

##q=RC\frac{dq}{dt}##
∴##\frac{1}{q}\frac{dq}{dt}=\frac{1}{RC}##
∴##\int_0^t \frac{1}{q}\frac{dq}{dt} dt=\int_0^t \frac{1}{RC} dt##
∴##\int_0^t \frac{1}{q} dq=\frac{t}{RC}##
∴##ln(q(t))-ln(q(0))=\frac{t}{RC}##
∴##ln(q(t))=\frac{t}{RC}##
∴##q(t)=e^{\frac{t}{RC}}##
∴##q(t)=C\epsilon(1-e^{-\frac{t}{RC}})##

This is the same result as the professor's solution, but with the correct values of R and C.

I hope this helps clarify any confusion and good luck with your studies!
 

1. What is an RC time circuit?

An RC time circuit is a circuit that consists of a resistor (R) and a capacitor (C) connected in series. It is commonly used in electronic devices to control the flow of current and create time delays.

2. How does the charge of a capacitor change over time in an RC time circuit?

The charge of a capacitor in an RC time circuit changes exponentially with time. At first, the capacitor charges rapidly, but as time goes on, the rate of charging decreases until it reaches its maximum charge.

3. What is the equation for the charge of a capacitor as a function of time in an RC time circuit?

The equation for the charge of a capacitor as a function of time in an RC time circuit is Q(t) = Q0(1 - e-t/RC), where Q0 is the initial charge of the capacitor, t is the time, R is the resistance, and C is the capacitance.

4. How do you derive the equation for the charge of a capacitor in an RC time circuit?

The equation for the charge of a capacitor in an RC time circuit can be derived using Kirchhoff's voltage law and Ohm's law. By analyzing the voltage across the resistor and the capacitor at different time intervals, the differential equation dQ/dt = -Q/RC can be obtained, which can then be solved to get the final equation.

5. What factors affect the charge of a capacitor in an RC time circuit?

The charge of a capacitor in an RC time circuit is affected by the initial charge of the capacitor, the resistance and capacitance values, and the time. It also depends on the voltage source connected to the circuit and any external factors that may affect the circuit's operation.

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