How Do You Calculate the Time Constant of an RL Circuit?

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SUMMARY

The time constant of an RL circuit can be calculated using the formula τ = R/L, where τ represents the time constant, R is resistance, and L is inductance. In this discussion, it is established that to reach 85% of the final value, the time taken is 1.86 seconds. By taking the natural logarithm of 0.15 (representing the 15% incomplete value), the number of time constants required is determined to be approximately 1.897. Dividing 1.86 seconds by 1.897 yields a time constant of approximately 0.980 seconds.

PREREQUISITES
  • Understanding of RL circuit theory
  • Familiarity with Kirchhoff's voltage law
  • Knowledge of natural logarithms and exponential functions
  • Basic skills in solving first-order ordinary differential equations (ODEs)
NEXT STEPS
  • Learn how to derive the time constant for RC circuits
  • Explore the implications of time constants in transient analysis
  • Study the behavior of RL circuits under different voltage sources
  • Practice plotting exponential functions to visualize circuit responses
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Electrical engineers, physics students, and anyone involved in circuit analysis or design will benefit from this discussion on calculating the time constant of RL circuits.

pmontone
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Can anyone tell me what formula I would use to find the time constant for a circuit that reaches 85% of its final value 1.86 seconds after the switch is closed.
 
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I am assuming that this circuit is in series with a constant voltage source, and initially there is no current when the switch is first thrown, for which case we have the first order ode arrising from taking Kirchhoff's voltage law around the loop.
[tex]\dot{I}+\frac{R}{L} I-\frac{\varepsilon_{0}}{R} = 0 , I(0) = 0 \cr<br /> I(t) = \frac{\varepsilon_{0}}{R} \left( 1- \exp(-t \tau) \right) , \tau = R/L[/tex]
where tau is the time constant you seek. Now use the information given, by taking the ratio @ t=0, and solve. Hope this helps, sincerely, x.
 
85% completion means 15% incomplete.

take the natural log of .15 (the decimal form of 15%). This is -1.897. toss out the negative as this is what you get when working with logs of numbers less than 1. what you are left with is 1.897. This is the number of time constants needed to reach 85% of completion.

divide you 1.86 seconds by 1.897, and you get about 0.980 seconds per time constant.

This is awfully close to 1, which might be the round answer you are expected to get. use this formula to plot 1,2,3,4,and 5 time constants, and you'll begin to see the pattern:

1-e-tc#.

that is: one minus (eulers number raised to the negative tc#). You'll get approximately .63, .87, .95, .98 and .99. But do it for yourself, with and without the "one minus"
 

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