A circuit consists of an inductor L and a resistor R connected in series to a battery via an open switch. After the switch is closed, how long must we wait for the current in the LR circuit to build up to 99.99% of its equilibrium value. Express the answer in terms of the number of “time constants,” L/R.Additional DetailsAnswer is 9.2 but i have no idea how to arrive at the answer :(
Instantaneous Current Maximum Current * ( 1 - e ^((-Time * R) / Inductance)) So just pick arbitrary, but useful values. If that formula is correct, then it should work for all values, however impractical. Right? 99.99 100 * ( 1 - e ^((-Time * 1) / 1)) Solve for Time, and you get 9.210 seconds, and since L/R 1, then 9.2 is also the number of time constants. I'll leave the trivial math up to you to write out. Try it with any values.
the present would be in section with the utilized voltage at resonance. that's while the capacitive reactance, Xc equals the Inductive reactance, XL. Xc XL a million/(2 x pi x f x C) 2 x pi x f x L f a million/(2 x pi x sq. Root of (L x C)) F 419 Hz. The resistor performs no area different than to limit the present at resonance to: utilized voltage /R and likewise to regulate 'Q'.
