In an oscillating LC circuit, L = 40.0 The difference of 12 V and a maximum ener

Question

In an oscillating LC circuit, L = 40.0 The difference of 12 V and a maximum energy of 270 mu J. At a certain instant the energy in the capacitor is 30 mu J. a) Find the maximum amount of charge that can be held by the capacitor. b) At that instant t = t_1, what is the potential difference across the capacitor? c) What is the current in the circuit as that instant t = t_1? d) How long does it take for the LC circuit from the moment of maximum capacitor to the moment of maximum energy in the inductor ? (Find the smallest time

Explanation / Answer

Given , LC osscilator,
L = 40 mH
Maxi8mum potential difference of capacitor, Vo = 12 V
maximum energy stored by capacitor = Eo = 270 micro J
at t= t1, Energy of capacitor, E = 30 micro J

a. Now, let capacitance of capacitor be C
then Maximum charge it can hold = Qo
and Qo = CVo also,
maximum energy = 0.5CVo^2 = Eo
so, C = 2Eo/Vo^2 = 2*30/12^2 = 0.41666 micro F
hence, Qo = 5 micro C

b. at t = t1, energy stored = 30 micro J
30*10^-6 = 0.5*C*V^2 ( where V is potential difference across capacitor at time t1)
V = 3.79 V
c. current in circuit at this instant = i
now, Ldi/dt = q/c = Lq"
let q = Qosin(wt)
then q" = -Qow^2sin(wt)
LQo*w^2sin(wt) = Qosin(wt)/C
w^2 = 1/LC
w = 1/sqroot(LC) = 7.348 rad/s

so, q = Qosin(wt)
i = Qowcos(wt)
now, energy at time t in capacitor = 0.5CV^2 = 0.5*q^2/C = 0.5*Qo^2sin^2(wt)/C = 30*10^-6
t = 0.21322 s
so, i = 0.14696*10^-6 A
d. for the energy to shift form one capacitor to other inductor, it takes quarter of time period
t = T/4
but T = 2*pi/w
so, t = 0.2137 s

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