Current using kirchhoff's loop rule

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SUMMARY

The discussion focuses on applying Kirchhoff's loop rule to derive the current \( I \) in an RL circuit, represented by the equation \( I = \frac{\epsilon}{R}\left(1 - e^{-\frac{Rt}{L}}\right) \). The key equation utilized is Kirchhoff's second rule: \( \epsilon - IR - L\frac{dI}{dt} = 0 \). Participants emphasize the importance of rearranging the equation correctly before integration, suggesting that isolating variables is crucial for successful integration. An alternative method proposed is to verify the solution by substituting it back into the differential equation.

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  • Understanding of Kirchhoff's loop rule and its application in circuit analysis.
  • Familiarity with differential equations, particularly first-order linear equations.
  • Knowledge of RL circuit behavior and the relationship between voltage, current, and inductance.
  • Basic integration techniques and variable separation methods.
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  • Study the derivation of the RL circuit response using differential equations.
  • Learn about the implications of Kirchhoff's laws in complex circuit analysis.
  • Explore numerical methods for solving differential equations in electrical circuits.
  • Investigate the behavior of RL circuits under different initial conditions and external voltages.
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Students studying electrical engineering, educators teaching circuit analysis, and anyone seeking to deepen their understanding of RL circuits and Kirchhoff's laws.

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Homework Statement


By applying kirchhoffs rules, show that the in the RL circuit is given by: [tex]I=\frac{\epsilon}{R}\left(1-e^{-\frac{Rt}{L}}\right)[/tex]


Homework Equations



kirchhoff's second rule: [tex]\epsilon-IR-L\frac{dI}{dt}=0[/tex]

The Attempt at a Solution


[tex]\epsilon=IR+L\frac{dI}{dt}[/tex]
[tex]{\epsilon}dt=IRdt+LdI[/tex]
[tex]\frac{\epsilon}{I}dt-Rdt=L\frac{1}{I}dI[/tex]
then i thought to integrate both sides, LHS w.r.t t and RHS w.r.t I, but doing this on paper doesn't get to the right answer, am i going wrong somewhere with the rearranging??
 
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You can't integrate it the way you have it written. You want to get all the I's on one side and all the t's on the other first, then you can integrate it.

You could also just plug the solution in and show it satisfies the differential equation instead of solving the equation.
 

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