# Finding R(t) in discharging RC circuit

• ddobre
In summary: The current is constant. When a capacitor C charged with Q is connected to a resistor R, current I will flow, and the capacitor voltage is Uc-RI=0. The capacitor voltage is Uc=Q/C and the current is defined as I=dQ/dt. The current is flowing off the capacitor now, so it decreases the charge on it.If it is constant, I =-I0, how does the charge change with time during the discharge?In summary, the current is constant, and the charge changes with time during the discharge.
ddobre

## Homework Statement

A charged capacitor with capacitance C is being discharged through a variable resistor that has its resistance dependent on time: R = R(t). Find function R(t) if the current through the resistor remains constant until the capacitor is completely discharged and the resistance at the initial moment of the discharge process (t = 0) is equal to R0

## Homework Equations

(1) I = (Q0/RC)e-t/RC
(2,3) Q0=Cε, Q = Cεe-t/RC
t = RC
IR = Q/C

## The Attempt at a Solution

Since I know I is contant, and at t = 0, R=R0, I tried to use equation (1) for R and at t = 0, when R = R0, so that I could equate the two equations and try to solve for R. This is what I started with:
Q0/R0C0 = (Q/RC)e-t/RC
I ended up with something like:
R = (QR0C0/Q0C)e-t/RC
But I was a little confused because there is still an R in the e expression. So I tried taking the natural log of each side, but what I ended up with didn't seem feasible. Any advice on what I should try to do?

Isn't your eqn (1) a solution to a differential equation which assumes R is constant (the solution, not the equation)?

haruspex said:
Isn't your eqn (1) a solution to a differential equation which assumes R is constant (the solution, not the equation)?

I think so. But I was just trying to define R in some way. I'm having trouble trying to find an equation for R(t)

ddobre said:
I think so. But I was just trying to define R in some way. I'm having trouble trying to find an equation for R(t)
See the first equation under https://en.m.wikipedia.org/wiki/Capacitor#DC_circuits
It is an integral equation, and it is obviously true. The equation just below it is obtained by differentiating it, but on the assumption that R is constant, so that second equation does not apply here.
Instead, you have that the current is constant.

When a capacitor C charged with Q is connected to a resistor R, current I will flow, and the capacitor voltage is Uc-RI=0. The capacitor voltage is Uc=Q/C and the current is defined as I=dQ/dt. The current is flowing off the capacitor now, so it decreases the charge on it.If it is constant, I =-I0, how does the charge change with time during the discharge?

Last edited:

## 1. How do I calculate the time constant (RC) of a discharging RC circuit?

The time constant (RC) of a discharging RC circuit can be calculated by multiplying the resistance (R) in ohms by the capacitance (C) in farads. This value represents the time it takes for the capacitor to discharge to 36.8% of its initial charge.

## 2. What is the equation for finding the voltage (V) at a specific time (t) in a discharging RC circuit?

The equation for finding the voltage (V) at a specific time (t) in a discharging RC circuit is V = V0e-t/RC, where V0 is the initial voltage on the capacitor.

## 3. Can I use the same equation to find R(t) in both charging and discharging RC circuits?

No, the equations for finding R(t) in charging and discharging RC circuits are different. For discharging circuits, the equation is R(t) = R0e-t/RC, where R0 is the initial resistance in the circuit.

## 4. How does the value of the capacitor affect the rate at which the voltage decreases in a discharging RC circuit?

The value of the capacitor directly affects the time constant (RC) of the circuit, which in turn determines the rate at which the voltage decreases. A larger capacitor will have a longer time constant, resulting in a slower decrease in voltage. Similarly, a smaller capacitor will have a shorter time constant and a faster decrease in voltage.

## 5. Is it possible to determine the exact time it takes for a capacitor to fully discharge in an RC circuit?

Technically, no. Theoretically, a capacitor will never fully discharge in an RC circuit, as it will always retain a small amount of charge. However, practically, the discharge can be considered complete when the voltage on the capacitor reaches 0. Therefore, the time it takes for the capacitor to reach 0 voltage can be calculated using the time constant (RC) equation.

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