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Is it the time when the current throught the inductor is 0.63 times of the maximum possible current through the inductor?

Or is it the time when the current is 0.63 times of the maximum possible current?

The difference between 1st's maximum current and 2nd's maximum current is... if inductor is present in parallel with source then the current will not be same... I mean current through the L and current from the source.

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  • \$\begingroup\$ 0.63 times of the DIFFERENCE between the PRESENT current and steady state current OF THE CIRCUIT \$\endgroup\$ – DKNguyen Nov 15 '20 at 20:15
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    \$\begingroup\$ When the system undergoes a state change where the setpoint is moved from one place to another, the process variable (current, perhaps) at the moment of that state change begins moving towards the new setpoint. If the behavior is of 1st order, then it will follow an exponential curve towards the new setpoint. When it reaches \$1-e^{-1}\approx 63.212\$% of the way, that's one time constant. Note the -1 as the power of e. That's when \$\frac{t}{\tau}=1\$. The concept applies to many circumstances and process variables. Not just inductor current. \$\endgroup\$ – jonk Nov 15 '20 at 20:24
  • \$\begingroup\$ @jonk or put another way the difference between the current value and the target is 36.7% after 1 time constant period, 0.367 * 0.367= 13.4% after 2, ... and so forth until it is at 99.5% of it's final value at 5 time constants \$\endgroup\$ – mhaselup Nov 16 '20 at 0:13
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Is it the time when the current through the inductor is 0.63 times of the maximum possible current through the inductor?

No. The "maximum possible current" is not right. It is the time taken for the current to rise 63% of the way from an initial value of current to the steady state current for a given applied voltage.

Or is it the time when the current is 0.63 times of the maximum possible current?

No. The "maximum possible current" is not right. It is the time taken for the current to rise 63% of the way from an initial value of current to the steady state current for a given applied voltage.

The difference between 1st's maximum i and 2nd's maximum i is... if inductor is present in parallel with souce then the current will not be same... I mean i through the L and current from source

This is jumbled gibberish so I don't know what you mean. Draw a diagram if you require a further answer. If English is not your first language then put that information in your user profile so we know that you may need assistance.

Another way of looking at it is shown below. (This is an RC charge curve but an RL curve would look the same for current rather than voltage.

enter image description here

Figure 1. When a tangent line is drawn at the beginning of a capacitor's charging curve, it will intercept the asymptote representing the capacitor's final charge at t = RC. Image source: Algebra Lab.

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  • \$\begingroup\$ The really cool thing is that if you plot the exponential approaches to steady state from different initial conditions (starting at the same time), the tangents will al intersecate in the same point. \$\endgroup\$ – Sredni Vashtar Nov 16 '20 at 16:24
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The time constant is when the initial slew_rate (the charging rate) would intercept the final steady_state value.

Thus a system with 1_pole response, moving from 0v to +5v, with 1uS timeconstant, would need 1 uS to move from 0v to 5v IF the slewrate did not change. Being the slewrate does change because the charging rate changes, because the voltage across the resistor is constantly becoming smaller, we get that popular exponential settling.

The (initial) slewrate is thus 5v/1uS = 5 volts per microsecond.

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  • \$\begingroup\$ Are you sure? OP asking about time constant, not slew rate. \$\endgroup\$ – relayman357 Nov 15 '20 at 21:05
  • \$\begingroup\$ Sorry, i'm a little slow. You are saying that a 1-pole circuit - like the current in an RL circuit at energization from a simple dc voltage source - will go from 0 to V/R in 1 time constant? Where am i missing the gist of what you and @analogsystemsrf are saying? Re: "...would need 1 uS to move from 0v to 5v" \$\endgroup\$ – relayman357 Nov 15 '20 at 22:16
  • \$\begingroup\$ @relayman357, I've added a diagram to my answer. See if it helps. For decades I didn't realise that the curve tangent crossed the fully-charged line at t = RC. \$\endgroup\$ – Transistor Nov 15 '20 at 22:36
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    \$\begingroup\$ This question is about time constants in physical systems (big clue is the 63% change). \$\endgroup\$ – mhaselup Nov 16 '20 at 0:04
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    \$\begingroup\$ His edit clears it up. \$\endgroup\$ – relayman357 Nov 16 '20 at 8:01

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