Question
Hey all, My question says that I have an 8.00 mH inductor, a4.00 resistor, and a 6.00V battery. They are allconnected in a series, and there is a switch. I have to find (1) the inductive time constant of the circuit(2) the current in the circuit .000200 sec after the switchis closed, (3) the value of the final steady-state current andfinally (4) the lenght of time it takes the current to reach 80.0%of its maximum value. Thanks for any and all help!!!! Landon Hey all, My question says that I have an 8.00 mH inductor, a4.00 resistor, and a 6.00V battery. They are allconnected in a series, and there is a switch. I have to find (1) the inductive time constant of the circuit(2) the current in the circuit .000200 sec after the switchis closed, (3) the value of the final steady-state current andfinally (4) the lenght of time it takes the current to reach 80.0%of its maximum value. Thanks for any and all help!!!! Landon
Explanation / Answer
(1) time constant = L / R where L = inductance = 8 m H = 0.008 H R= resiatnce = 4 So, = 0.002 s (2). current after 0.002 s after switch closed i = I [1- e -t / ] where I = maximum current = maximum voltage /resistance = 6 V / 4 = 1.5 A substitue values weget i = 0.945 A (3). final steady current I = 1.5 A (4). let the required time be t we know i = I [ 1- e -t / ] given i = 80 % of I = 0.8 I substitue we get 0.8 = [ 1- e -t / 0.002] e -t / 0.002 = 0.2 -t / 0.002 = -1.609 t = 3.2188 ms