Calculating Current in a Single-Loop Circuit with Resistor and Capacitor

[reising1]

Homework Statement

There is a Battery connected to a single loop circuit containing two resistors, R1 and R2, and one capacitor L.

After a long time, the battery is removed, so there is a single loop circuit with just two resistors and a capacitor.

What is the current going through R1?

The Attempt at a Solution

This is what I thought:

So, the EMF is removed. Thus, using a loop rule, we have the formula:

0 = IR + L(di/dt) where R is R1+R2

Integrating, we have

0 = (1/2)(I^2)(R) + (L)(I)

Dividing everything by I, we have

0 = (1/2)(I)(R) + L

Thus, I = (2L) / R

However, this is incorrect.

Any ideas?

Thanks!

 

[LeonhardEuler]

There are a few problems here. You call this an RL circuit, and you use the equations for an RL circuit, but you say it contains a capacitor instead of an inductor. I assume you are just using the wrong word.

The second problem is here:

0 = IR + L(di/dt) where R is R1+R2

Integrating, we have

0 = (1/2)(I^2)(R) + (L)(I)

Are you sure you did that correctly? What variable are you integrating with respect to?

 

[reising1]

Oh, wait. That is wrong.

With respect to I, you get:

R + LI = 0

So I = -R/L?

 

[reising1]

Deriving with respect to I. Sorry.

 

[LeonhardEuler]

This is still not right because you differentiated one term and integrated the other.

 

[LeonhardEuler]

You might want to rethink the whole strategy of trying to take an integral or a derivative.
You have both I and dI/dt in this equation. What kind of equation does that make it?

 

[reising1]

What do you mean?

d/di(RI) = R
d/di(L(di/dt)) = LI

 

[reising1]

differential equation

 

[LeonhardEuler]

What do you mean?

d/di(RI) = R
d/di(L(di/dt)) = LI

Taking a derivative of a derivative doesn't make the derivative do away.

 

[LeonhardEuler]

differential equation

Yes!