Question

3.A 12V battery is in series with a 10ohm resistor ,an open switch and a 20mH inductor. a) What is the current flowing in the circuit after the switch has been closed for a long time? b) What is the value of time constant c)How much current will be flowing in the circuit after 10ms d) What is the voltage across the resistor and inductor at t=10ms e) At what rate is the current changing at this instant f)How many time constants will it take for the current to reach 99.4% to maximum value​

Answer 1

A. I = 12/(10 + 20mH) = 0.545 A

B. Time constant = 20mH

C. I = 0.545 * (1 - e^(-10/20mH)) = 0.49 A

D. V = 12 - IR = 12 - (0.545 * 10) = 10.91 V

E. di/dt = 12/20mH = 0.6 A/s

F. It will take 3 time constants for the current to reach 99.4% of its maximum value.

Step-by-step explanation: