A 2.50mH inductor is connected in series with a dc battery of negligible internal resistance, a 0.760kOhm resistor, and an open switch.How long after the switch is closed will it take for the current in the circuit to reach half of its maximum value ?How long after the switch is closed will it take for the energy stored in the inductor to reach half of its maximum value?
L 2.5*10^-3 H R 760 Ω This will produce a very short time constant T L / R 3.2895 * 10^-6 s The current is i Imax * (1 - e^(-t/T)) When i 0.5*Imax 0.5 1 - e^(-t/T) e^(-t/T) 1 - 0.5 0.5 -t / T ln(0.5) -0.69315 t --0.69315 * 3.2895*10^-6 s 2.28*10^-6 s 2.28 μs Emax 0.5*L*Imax^2 When E 0.5*Emax i^2 / Imax^2 0.5 so i Imax / sqrt(2) 0.7071*Imax 0.7071 1 - e^(-t/T) e^(-t/T) 1 - 0.7071 0.2929 -t / T ln(0.2929) -1.228 t --1.228 * 3.2895*10^-6 s 4.04*10^-6 s 4.04 μs
