Bac S physique 2008 Exercice II : Un réveil en... In uence d'une bobine dans un circuit électrique

R1=r
τ=L
R
[R] = M∗L2∗T−3∗I−2
[L] = M∗L2∗T−2∗I−2

u=R∗i⇒[u]=[R]∗I
u=L∗di
dt ⇒[u]=[L]∗I∗T−1
T−1[u] = [L]∗I∗T−1= [R]∗I⇒[τ] = [L]
[R]=T
τ L = 1H R = 10Ω τ= 0,1s
5∗τ= 0,5s
T= 30ms
T= 2 ∗Π∗√LC ⇒L=1
C∗(T 2∗Π)2
L=1
22∗10−6∗(30∗10−3 2∗Π)2= 1.036H L = 1.0H

λ= 400 800nm
λ≥800nm

UR0
UBD
Y1Y2
R0
UBD R0UR0
UBD =UR0+UBC u(t) = UBC =UBD −UR0
i(t)
R0UR0=
R0∗i(t)⇒i(t) = UR0
R0
p(t) = u(t)∗i(t) = (UBD −UR0)∗UR0
R0
R0
R0
0.90 ∗11.2 = 10.08W