An electrical engineer designed an RL circuit similar to the one shown in the figure, with E = 6.00 V, L = 5.00 mH, and R = 8.00 0.R(a) What is the inductive time constant of the circuit (in ms)?.625ms(b) Calculate the current in the circuit (in A) 250 µs after the switch is closed.Apply the equation for the current as a function of time in an RL circuit, and use your time constant calculated in part (a). Be sure your time units are consistent. A(c) What is the value of the final steady-state current (in A)?Apply Ohm's law. A(d) After what time interval (in ms) does the current reach 80.0% of its maximum value?msNeed Help?Read It

Hello. Since your question has multiple sub-parts, we will solve the first three sub-parts for you. If you want remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve. (a) Let L denotes the inductance, R denotes the resistance, and T denotes the inductive time constant. Therefore, Let t denotes the time, V denotes the battery voltage, and I denotes the current. Therefore, the current in series LR circuit can be represented as, Let I0 denotes the current after a long time. Since the inductor act as a conducting wire after a long time, current can be obtained as,

Answered: An electrical engineer designed an RL…