Question
Consider the circuit above, the switch has been opened for along time ( assume the inductor is ideal , that is it has noresistance)
a) What is the magnitude of the the current through theinductor in the instant after the switch is closed. show allcalculations
b) What is the magnitude of the current, increasing, ordecreasing as a function of time?
Suppose the switch S is close and remained closed for a very longtime:
c) what is the magnitude of the current through the inductor afteryou wait a very long time.
d) Suppose the switch S is close and remained closed for a verylong time what is the magnitude of the current constant, increaseinor decreasing as the function of time.
Explanation / Answer
a)the magnitude of the the current through the inductor in theinstant after the switch is closed is I = (E/R) * (1 - e-Rt/L) = (E/R) * (1 -e-t/L/R) = (E/R) * (1 -e-t/) where the constant is the time constant of the RLcircuit Here,t = 0 or I = (E/R) * (1 - e0) = 0 b)The magnitude of the current is decreasing as a function of time.This is because when the switch is closedinstantly then we consider the time as t = 0. c)the magnitude of the current through the inductor after youwait a very long time is I = (E/R) * (1 - e-Rt/L) Here,t = or I = (E/R) * (1 - e-R()/L) = (E/R) * (1 -e-) = (E/R) * (1 - (1/e)) =(E/R) * (1 - (1/)) = (E/R) or I = (E/R) = (12/45) = 0.27 ampere d)the current flowing through the circuit is I = (E/R) * (1 - e-Rt/L) or I = Io* (1 - e-t/) solving the above equation for = [t/ln(Io/Io - I)] when the switch S is closed for a long time then t = therefore = [/ln(Io/Io - I)] = therefore,the time constant of the circuit increases as afunction of time. = [/ln(Io/Io - I)] = therefore,the time constant of the circuit increases as afunction of time.