An L-C circuit consists of a 69.5-mH inductor and a 240-µF capacitor. The initial charge on the capacitor is 5.95 µC, and the initial current in the inductor is zero.
(a) What is the maximum voltage across the capacitor?
__________ V
(b) What is the maximum current in the inductor?
__________ A
(c) What is the maximum energy stored in the inductor?
__________ J
(d) When the current in the inductor has half its maximum value,
what is the charge on the capacitor?
__________C
When the current in the inductor has half its maximum value, what
is the energy stored in the inductor?
_____________ J
A.
Q = C*V
Vmax = Q/C = 5.95*10^-6/(240*10^-6)
Vmax = 0.0247 V
Part B
Energy stored is given by:
U = 0.5*L*i)max^2 = 0.5*Q^2/C
i)max = Q/sqrt (LC)
i)max = 5.95*10^-6/sqrt (69.5*10^-3*240*10^-6)
i)max = 0.00146 Amp = 1.46*10^-3 Amp
Part C
Umax = 0.5*L*i)max^2
Umax = 0.5*69.5*10^-3*(1.46*10^-3)^2 = 7.41*10^-8 J
Part D
If i = i)max/2, then energy stored in inductor
UL = Umax/4
UC = 3*Umax/4
UC = 3*Q^2/(4*2*C) = q^2/2C
q = Q*sqrt (3/4)
q = 5.95*10^-6*sqrt (3/4) = 5.15*10^-6 C
Part E.
energy stored in inductor will be
UL = Umax/4
UL = 7.41*10^-8/4 = 1.85*10^-8 J
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