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RL circuit 
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#1
Apr310, 04:22 PM

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 


#2
Apr310, 05:20 PM

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: 


#3
Apr310, 05:34 PM

P: 55

Oh, wait. That is wrong.
With respect to I, you get: R + LI = 0 So I = R/L? 


#4
Apr310, 05:34 PM

P: 55

RL circuit
Deriving with respect to I. Sorry.



#5
Apr310, 05:36 PM

PF Gold
P: 864

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



#6
Apr310, 05:38 PM

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? 


#7
Apr310, 05:38 PM

P: 55

What do you mean?
d/di(RI) = R d/di(L(di/dt)) = LI 


#8
Apr310, 05:38 PM

P: 55

differential equation



#9
Apr310, 05:39 PM

PF Gold
P: 864




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