An LR combination is connected to an idal battery. If l = 10 mH, R = 2.0 Omega and epsilon = 2.0 V, how much time will it take for the current ot reach 0.63 A?

Explanation

The steady state current in the LR circuit is `i_0 = epsilon/ R = 2
0 V / 2
0 Omega = 1 A`
Thus, 0
63 A is 63% of the steady- state current
As we know, it takes one time constant for the current to reach 63% of its steady-state value
Hence the required time = L/R = 10 mH / 2
0 Omega = 5
0 ms