A solenoid of inductance 50 mH and resistance 10 Omega is connected to a battery of 6V. Find the time elapsed before the current acquires half of its steady - stae value.

Explanation

The time constant of the circuit is `tau = L/R = 50 mh/10 Omega = 5 ms`
The current at tiem t is given by `I = i_0(1-e^(-t/tau))`
For `I = i_0/2, i_0/2 = i_0(1-e^(-t/tau))` or, `e^(-t/tau) =(1)/(2) or, (t)/(tau) = In2` giving t = tau In 2 = (5ms)(0
693) = 3
5 ms