RL circuit

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!

The Attempt at a Solution

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LeonhardEuler
Gold Member
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:
reising1 said:
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?

Oh, wait. That is wrong.

With respect to I, you get:

R + LI = 0

So I = -R/L?

Deriving with respect to I. Sorry.

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

LeonhardEuler
Gold Member
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?

What do you mean?

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

differential equation

LeonhardEuler
Gold Member
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
Gold Member
differential equation
Yes!