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.