a headlight on bike consists of a light bulb with 1.00?? resistance and a 9.00V

Question

a headlight on bike consists of a light bulb with 1.00?? resistance and a 9.00V battery. Pat would like the light to remain on for a short time after she parks the bike and opens the switch so she can see the path to the door. the light is still visibel as long as the current is above 450.0mA. she makes an inductor with very low resistance wire wound around a 1.00 cm diameter pen cap that is 3.00 cm long. she measures the inductance with a multimeter and gets a value of 20.0 H. assuming the light has been closed for some time ans using this inductor,

a) what is the time constant for the circuit, t?

b) how many turns did pat wind the wire about the cap/

c) how much energy is stored in the inductor before the switch is opened?

d)how long is the light lit after the switch is opened?

e)what is the magnetic flux through the inductor 2.00s after the switch is opened?

Explanation / Answer

(a) T = L/R = 20 / 1 = 20 sec

(B) L = u0 N^2 A / l

20 = (4pi x 10^-7) (N^2) (pi x 0.005^2) / (0.03)

N = 77970 turns

(C) i = 9 / 1 = 9 A

energy store = Li^2 / 2 = 20 x 9^2 / 2 = 810 J

(D) I = Io [ e^(-t/T)]

0.450 = 9 [ e^(-t / 20)]

t = 60 sec

(e) I = 9 [ e^(-2/20)] = 8.14 A

flux = L i = 162.8 Wb

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.