Question

/000t 10A ine V ). 120.2 50A HOmH

Please provide me with the correct solution, step by step.

Answer 1

At t = 0: - Inductor is short circuited. By nodal analysis v(0^bar)/40 + V(0^bar)/8 + V(0^bar)/10 = 50 implies v(0^bar) + 5 v(0^bar) + 4 v(0^bar) = 2000 implies v(0^bar) = 200 volts tau(for t > 0) tau = L/R_th R_th: - Independent voltage source is short circuited 120//40 = 120 times 40/160 = 30 ohm R_th = 10 + 30 = 40 ohm tau = L/R_th = 40m/40 = 0.001 sec At t = infinity Inductor is short circuited By nodal analysis v(infinity)/40 + V(infinity)/10 + V(infinity) - 800/120 = 0 implies 3 v(infinity) + 12 v(infinity) + v(infinity) - 800 = 0 implies 16 v(infinity) = 800 implies v(infinity) = 50 volts we know that, v(t) = v(infinity) + [v(0^bar) - v(infinity)]e^-t/tau = 50 + [200 - 50]e^-t/0.001 v(t) = 50 + 150 e^-1000t volts This is the correct solution.