An L-C circuit consists of a 58.0-mH inductor and a 255-plF capacitor. The initi

Question

An L-C circuit consists of a 58.0-mH inductor and a 255-plF capacitor. The initial charge on the capacitor is 6.25 ?C, and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (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?

Explanation / Answer

part a:

maximum voltage across capacitor=initial charge/ capacitance

=6.25*10^(-6)/(255*10^(-6))

=0.024510 volts

part b:

maximum energy in the capacitor=0.5*capacitance*voltage^2

=0.5*255*10^(-6)*0.02451^2

=7.6594*10^(-8) J

let maximum current in the inductor is I A.

then maximum energy=0.5*inductance*I^2

==>0.5*58*0.001*I^2=7.6594*10^(-8)

==>I=0.0016252 A

part c:

maximum energy stored in the inductor=7.6594*10^(-8) J

part d:

when the current in the inductor has half its maximum value,

current=I/2=8.1259*10^(-4) A

energy in the inductor=0.5*L*current^2

=0.5*58*0.001*(8.1259*10^(-4))^2

=1.9149*10^(-8) J

energy in the capacitor=total energy in the system - energy in the inductor

=7.6594*10^(-8) - 1.9149*10^(-8)

=5.7445*10^(-8) J

then if charge on the capacitor Q,

Q^2/(2*C)=5.7445*10^(-8)

==>Q=sqrt(2*255*10^(-6)*5.7445*10^(-8))=5.4127*10^(-6) C

part e:

when the current in the inductor has half its maximum value, energy stored in the inductor=1.9149*10^(-8) J (from part d)

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