The current in a series RL circuit

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In a series RL circuit, the current reaches 20% of its final value in 3.4μs with an inductance of 2.0mH. The equation I=I_0(1-e^(-t/(L/R)) is questioned for its accuracy in solving for resistance R. Participants express uncertainty about how to apply the equation and determine the final current value and the current after 3.4μs. Clarification is sought on the correct approach to find R using the given parameters. Understanding the relationship between time, inductance, and resistance is crucial for solving the problem.
BryceHarper
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Homework Statement



The current in a series RL circuit increases to 20% of its final value in 3.4μs. If L = 2.0mH, what is the resistance R?

Homework Equations



I=I_0(1-e^(-t/(L/R))) <----Im not sure if this equation is right?


The Attempt at a Solution



I'm unsure how to start this problem, or if I'm on the right track with this equation?
 
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BryceHarper said:
The current in a series RL circuit increases to 20% of its final value in 3.4μs. If L = 2.0mH, what is the resistance R?
I=I_0(1-e^(-t/(L/R))) <----Im not sure if this equation is right?
Neither do I, but assuming it is...
In terms of the variables in that equation, what would be the final value, and what would be the value after 3.4μs?
 

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