An emf of 15 V is applied to a circuit containing 5 H inductance and 10 Ω resistance. The ratio of currents at time t = ∞ and t = 1 s after closing the switch is:

Step-by-step solution

Given: The inductance L = 5   H , resistance R = 10   Ω and emf applied V = 15   V . The time constant for L R constant is given by, τ L = L R = 1 2   s The current in the L R circuit is given by, i = i 0 1 - e - t / τ L ; where i 0 = V R is the maximum current at t = ∞ . The ratio of the maximum current to current will be, ∴   i 0 i = 1 1 - e - t / τ L Substituting, t = 1   s and τ L = 1 2   s We get, i 0 i = e 2 e 2 - 1 .