A series LR circuit consists of a 2.0-H inductor with negligible internal resistance, a 100-ohm resistor, an open switch, and a 9.0-V ideal power source. (a) If the switch is closed at time t=0s, what is the current 5.8ms later? (b) After the switch is closed, what is the maximum power delivered by the power supply?

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As per the theory of growth of current in LR circuit , current at time t  I(t) = (V/R)(1-e-Rt/L) ...(1)  Here V = emf of battery =9.0V , R== resistance of resistor=100Ω,L=inductance of inductor=2H  a)Put t= 5.8ms=0.0058s  Substitute numerical values I(t)=(9.0V/100Ω){1-e-(100Ω)(0.0058s)/(2H)} = 0.023 A. b) maximum power delivered by power supply  Pmax = V imax  From (1) imax = V/R  Then ,Pmax =V(V/R)= V²/R =( 9V)²/100Ω=0.81W   ...