Time taken for the current to fall?

Question

A 2H coil of neligible resistance while connected to a power supply carries 5A. If it is dissconnected from power supply and connected to a 120 ohms discharge resistor,calculate the time taken for the current to fall to 0.02 seconds

Answer 1

If an inductance L carrying a current I is discharged through a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vL across the inductance and the magnetic linkage yL in the inductance are:

i = Ie - tR / L

vR = iR = IRe - tR / L

vL = vR = IRe - tR / L

yL = Li = LIe - tR / L

Rise Time and Fall Time

The rise time (or fall time) of a change is defined as the transition time between the 10% and 90% levels of the total change, so for an exponential rise (or fall) of time constant T, the rise time (or fall time) t10-90 is:

t10-90 = (ln0.9 - ln0.1)T » 2.2T

The half time of a change is defined as the transition time between the initial and 50% levels of the total change, so for an exponential change of time constant T, the half time t50 is :

t50 = (ln1.0 - ln0.5)T » 0.69T

Note that for an exponential change of time constant T:

- over time interval T, a rise changes by a factor 1 - e -1 (» 0.63) of the remaining change,

- over time interval T, a fall changes by a factor e -1 (» 0.37) of the remaining change,

- after time interval 3T, less than 5% of the total change remains,

- after time interval 5T, less than 1% of the total change remains.