The RL circuit has a 12 V battery and a 25 ohm resistor The

The RL circuit has a 12 V battery and a 25 ohm resistor. The time constant for the circuit is 3.8 X 10(-3) seconds. A)calculate the inductance of this circuit b) find the current in the circuit a long time after the switch was closed C) at what time is the current 0.2 A

Solution

emf of the battery E = 12 volt

Resistance R = 25 ohm

Time constant T = 3.8 x10 -3 s

We know T = L/R

(A). Inductance of the circuit L = RT

                                            = 25(3.8 x10 ^-3)

                                            = 95 x10 ^-3 H

(B).the current in the circuit a long time after the switch was closed io = V/R

             io=12/25

               = 0.48 A

C) the current i = 0.2 A

We know in LR circuit i = io [1-e ^-t/T]

                             0.2 = 0.48 [1- e^-t/T]

                   1-e^-t/T = 0.2/0.48

                                  = 0.41666

                         e^-t/T = 1-0.41666

                               = 0.58333

                        -t/T = ln(0.58333)

                             = -0.5389

                           t = 0.5389 T

                            = 0.5389(3.8 x10 ^-3 s)

                            = 2.048 x10 ^-3 s