Time taken for current?

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 0.025 A

Answer 1

The inductor has stored energy in it's magnetic field.  This is an exponential decay.

Time constant τ = L/R = 2H/120Ω = 0.0167s = 16.7ms

The exponential decay function gets smaller and smaller but never gets to 0, so the inductor is considered fully discharged (by electrical engineers) after 5 time constants or 5 * 16.7ms = 83.5ms, when it has a value of i = 1% of I_I = 1% of 5A = 50mA.  So you want  i = 25mA.

i = I_I e^ (-t/τ)  -- solve for t.

i/I_I = e^ (-t/τ)

take natural log of both sides.

-t/τ = ln (i/I_I)

t = -τ ln (i/I_I) = - 16.7ms ln (25mA/5A) = 88.5ms.

where

τ = (Greek letter tau) Time constant in seconds (s)

i = instantaneous current in amps (A).

t = time in seconds (s).

I_I = Initial current in amps (A) = 5A