LR Circuit: Solving for Charge Flow?

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SUMMARY

The discussion centers on solving for charge flow in an LR circuit, specifically using the equation for current, i = i_o (1 - e^{-t/\tau}), where τ = L/R. The user expressed difficulty in integrating the equation dq/dt = i_o (1 - e^{-t/\tau}) to find the charge q. The solution involves integrating with respect to time t to derive the correct expression for charge flow.

PREREQUISITES
  • Understanding of LR circuits and their behavior
  • Familiarity with differential equations
  • Knowledge of integration techniques
  • Basic concepts of electrical engineering
NEXT STEPS
  • Study the integration of exponential functions in the context of electrical circuits
  • Learn about the time constant τ in RL circuits and its implications
  • Explore the relationship between current and charge in electrical circuits
  • Review examples of charge flow calculations in LR circuits
USEFUL FOR

Students and professionals in electrical engineering, particularly those studying circuit analysis and dynamics in RL circuits.

cupid.callin
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I solved it :p

Homework Statement


attachment.php?attachmentid=33120&stc=1&d=1300218625.jpg

The Attempt at a Solution



I need help with part (a)

Current in LR circuit, [tex]i = i_o (1 - e^{-t/\tau})[/tex] and [tex]\tau = \frac{L}{R}[/tex]

so, [tex]\frac{dq}{dt} = i_0 (1 - e^{-t/\tau})[/tex]

integrating it is not giving me the correct answer.
 

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cupid.callin said:
I solved it :p


Homework Statement


attachment.php?attachmentid=33120&stc=1&d=1300218625.jpg



The Attempt at a Solution



I need help with part (a)

Current in LR circuit, [tex]i = i_o (1 - e^{-t/\tau})[/tex] and [tex]\tau = \frac{L}{R}[/tex]

so, [tex]\frac{dq}{dt} = i_0 (1 - e^{-t/\tau})[/tex]

integrating it is not giving me the correct answer.

just integrate with respect to t, dq/dt = ... dq = ...dt
 
its solved ... i wrote it on the top
 

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