A 0.38-H inductor and a 211.0-? resistor are connected in series to a 9.0-V batt

Question

A 0.38-H inductor and a 211.0-? resistor are connected in series to a 9.0-V battery.

(a) What is the maximum current that flows in the circuit?
0.043 A

(b) How long after connecting the battery does the current reach half its maximum value?
1.25e-3 s

(c) When the current is half its maximum value, find the energy stored in the inductor.
8.6e-5 J

When the current is half its maximum value, find the rate at which energy is being stored in the inductor.
_____________ W

When the current is half its maximum value, find the rate at which energy is dissipated in the resistor.
____________ W

(d) Redo parts (a) and (b) if, instead of being negligibly small, the internal resistances of the inductor and battery are 75 ? and 20.0?,respectively.
What is the maximum current that flows in the circuit?
2.9e-2 A

How long after connecting the battery does the current reach half its maximum value?
8.6e-4 s

Explanation / Answer

A) Imax = V/R= 9/211 = 0.043 A

B) I = Imax*e^(-t/T)

T = L/R = 0.38/211 = 0.0018 S

0.5*Imax = Imax*e^(-t/0.0018)

0.5 = e^(-t/0.0018)

-t/0.0018 = ln(0.5)

t = 1.25*10^-3 S

C) U = 0.5*L*I^2 = 0.5*0.38*(0.043/2)^2 = 8.6*10^-5 J

rate at which energy dissipated by the resistor is P = I^2*R = (0.043/2)^2*211 = 0.097 W

rate at which Energy delivered by the battery P = V*I = 9*(0.043/2)= 0.1935 W

then rate at which energy is stord in the inductor is 0.1935-0.097 = 0.0965 W

D) Imax = 9/(211+75+20) = 2.9*10^-2 A

0.5 = e^(-t/T)

T = L/R = (0.38)/(211+75+20) = 1.24*10^-3 S

-t = ln(0.5)*1.24*10^-3

t = 8.6*10^-4 S

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