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## 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!