# RL circuit

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 P: 55 1. The problem statement, all variables and given/known data 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? 3. 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! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
PF Gold
P: 864
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:
 Quote by reising1 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?
 P: 55 Oh, wait. That is wrong. With respect to I, you get: R + LI = 0 So I = -R/L?
 P: 55 RL circuit Deriving with respect to I. Sorry.
 PF Gold P: 864 This is still not right because you differentiated one term and integrated the other.
 PF Gold P: 864 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?
 P: 55 What do you mean? d/di(RI) = R d/di(L(di/dt)) = LI
 P: 55 differential equation
PF Gold
P: 864
 Quote by reising1 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.
PF Gold
P: 864
 Quote by reising1 differential equation
Yes!

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