After having been open for a long time, the switch S was closed at time t = 0. T

Question

After having been open for a long time, the switch S was closed at time t = 0. The EMF of the ideal battery is 11.50 V, R = 7.30 Ohms, and L = 5.40 H.

After having been open for a long time, the switch S was closed at time t = 0. The EMF of the ideal battery is 11.50 V, R = 7.30 Ohms, and L = 5.40 H. At what rate is energy being delivered by the battery at time t = 1.050 s? How much energy has been delivered by the battery during the first 2.100 s after the switch has been closed? At what rate is energy being stored in the magnetic field of the inductor at time t = 1.050 s? How much energy has been stored in the magnetic field of the inductor during the first 2.100 s after the switch has been closed? At what rate is thermal energy being produced in the resistor at time t = 1.050 s? How much thermal energy has been dissipated in the resistor during the first 2.100 s after the switch has been closed?

Explanation / Answer

i = I0(1-exp(-t/tau), tau = L/R = 0.74, I0= V/R = 11.5/7.3 = 1.57

a)i = I0(1-exp(-t/tau)) = 1.57*(1-exp(-1.05/0.74)) = 1.19 A

b)i = I0(1-exp(-t/tau)) = 1.57*(1-exp(-2.1/0.74)) = 1.47 A

E = VIt = 11.5*1.47*2.1 = 35.50 J

c)E = 1/2Li^2 = 0.5*L*I0^2*(1-exp(-t/tau)^2

==> P = dE/dt = 2*0.5*L*I0^2*(1-exp(-t/tau))*exp(-t/tau)/tau

==> P = 2*0.5*5.4*1.57^2*(1-exp(-1.05/0.74))*exp(-1.05/0.74)/0.74 = 3.29W

d) E = 1/2Li^2= 0.5*L*I0^2*(1-exp(-t/tau)^2

==> E =  0.5*5.4*1.57^2*(1-exp(-2.1/0.74))^2 = 5.89 J

e)at t =1.05 i = 1.19

so H = i^2R = 1.19^2*7.3 = 10.33 W

f)at t = 2.1 i = 1.47

so H i^2R = 1.47^2*7.3 = 15.77 W

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