A coil of inductance 1 H and resistance 100 Ω is connected to a battery of 6 V . Determine approximately : (a) The time elapsed before the current acquires half of its steady-state value (b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln 2 = 0 . 693 , e - 3 2 = 0 . 25 )

Step-by-step solution

Given circuit is R - L growth circuit The current is the circuit is given by, i = E R 1 - e - t τ For current to be half of the peak value, i = E 2 R = E R 1 - e - t τ Solving t = τ ln 2 t = L R ln 2 = 1 100 0 . 693 = 0 . 00693 ≃ 7   ms At time t = 15   ms , i 15   ms = E R 1 - e - 15 10 ⇒ i = 6 100 1 - 1 4 = 3 4 × 6 100 = 0 . 045   A ⇒ U = 1 2 L i 2 by solving we get U = 1   mJ .