The switch in the circuit shown is closed at time t = 0. At what rate is energy

Question

The switch in the circuit shown is closed at time t = 0. At what rate is energy being dissipated as Joule heat in the resistor after an elapsed time equal to the time constant of the circuit? Answer in units of W Calculate the rate at which energy is being stored in the inductor at this time. Answer in units of W What is the total energy stored in the inductor at this time? Answer in units of J How long a time does it take the current to reach 89 percent of its maximum value? Answer in units of s

Explanation / Answer

30-4i=Vl Vl=-3i' => 30-4i=-3i` di/(7.5-i) =-4/3 dt ln(7.5-i) =-4/3 t+c =>i=7.5(1-Ae^(-4/3 t)) at t=0 i=0 =>A=1 i=7.5(1-e^(-1.33t)) A a)P=i^2 R = 7.5^2*4 (1-e^(-1.33t))^2 P(t=1/1.33) =7.5^2*4(1-e^(-1))^2=89.9047 W b)E=.5Li^2=3*.5*7.5^2 (1-e^(-1.33t))^2 dE/dt = .5*2*7.5^2*1.33 (1-e^(-1.33t))e^(-1.33t) E'(t=1/1.33) =7.5^2*1.33*(1-e^-1)e^-1 =17.3972 W c) E(t=1/1.33)=3*.5*7.5^2 (1-e^-1)^2 =33.7143 J d) i=7.5(1-e^(-1.33t)) A i=0.89*7.5 .89*7.5=7.5(1-e^(-1.33t)) =>.89=1-e^(-1.33t) =>e^(-1.33t)=.11 =>t=-ln(.11)/1.33 =1.65961 sec

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