A 12.0 V battery is connected into a series circuit containing a 14.0 N resistor and a 2.90 H inductor.(a) In what time interval (in s) will the current reach 50.0% of its final value?(b) In what time interval (in s) will the current reach 90.0% of its final value?What If? After a very long time, using a switch like that shown in the figure below, the battery is removed and the inductor is connected directly across the resistor.S2a오b(c) In what time interval (in s) will the current decrease to 50.0% of its initial value?(d) In what time interval (in s) will the current decrease to 10.0% of its initial value?

Given Potential difference = 12 V Resistance = 14 ohm Inductance = 2.90 H (a) R=14.0 Ω , L = 2.90 Hε = Imax 1-e-RtL 0.5 =-e-RtL ⇒In0.5= -RtL⇒ 0.693= 14 t2.90⇒t =0.143 s (b) ε = Imax 1-e-RtL0.9 = 1-e-RtL⇒e-RtL = 0.1 ⇒ln 0.1 = -RtL⇒2.3 = 14 t2.90⇒t =0.4764 s (c) I = Imax eRtL ln 0.5 = 1 eRtL⇒0.693 = 14t2.9⇒ t=0.143 s (d) 0.1 = 1-e-RtL⇒e-RtL = 0.1 ⇒ln 0.1 = -RtL⇒2.3 = 14 t2.90⇒t =0.4764 s

Answered: A 12.0 V battery is connected into a…