Question

The current in an RL circuit builds up to one-third of its steady-state value in 4.36 s. Find the inductive time constant.

Answer 1

In RL circuit, we have

i = (V / R) [1 - e^-t/_L] eq.1 }

where, i = steady state current

if the current is one-third of the steady state value, then at given time 4.36 sec -

(V / 3R) = (V / R) [1 - e^-t/_L]

e^-t/_L = [1 - (1/3)]

e^-t/_L = (2 / 3)

^-t/_L = log (2 / 3)

^-t/_L = - (0.4054)

_L = t / (0.4054)                                                                      { eq.2 }

inserting the value of 't' in eq.2,

_L = (4.36 sec) / (0.4054)

_L = 10.75 sec

The inductive time constant will be given as, _L= 10.75 sec