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If you have a huge time constant in LR circuits, does this mean it takes a long time to discharge and lose all current in circuit?

Similarly, if you have a tiny time constant in LR circuits, does this mean it takes a very short time to discharge circuit?

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  • $\begingroup$ LR circuits do not contain capacitors. $\endgroup$ – sammy gerbil Mar 18 '17 at 19:00
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    $\begingroup$ Apologies! I meant, discharge & lose all current in circuit? $\endgroup$ – Marie R Mar 18 '17 at 19:02
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    $\begingroup$ Yes, that's it. More info en.wikipedia.org/wiki/Time_constant $\endgroup$ – JMLCarter Mar 18 '17 at 19:08
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You are correct. The larger the time constant, the longer it takes for current to build up or decay in LR circuits, or for capacitors to charge or discharge in CR circuits.

Same with half-life of radioactive materials, which is a kind of time constant. Longer half-life means it takes longer to decay.

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  • $\begingroup$ This is not necessarily true for the decay of current. If you open a switch in an LR circuit, the current doesn't keep flowing just because the time constant is long! $\endgroup$ – alephzero Mar 19 '17 at 0:49
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    $\begingroup$ I am sorry, I don't see the relevance of your comment to my answer. $\endgroup$ – sammy gerbil Mar 19 '17 at 0:59
  • $\begingroup$ @alephzero would you please explain? You mean the current would stop instantly? $\endgroup$ – thermomagnetic condensed boson Nov 3 '18 at 12:19
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One time constant specifies how long it takes to see 63% of the total change, as given by $1-e^{-1}$. This means that a long time constant requires a long time to see 63% of the total change, while a short time constant requires a short time to see 63% of the total change. Note that the time-constant concept is used because the exponential decay function never quite gets to zero mathematically.

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