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Alternating Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current5 Marks
Question
If a $LC$ circuit is considered analogous to a harmonically oscillating spring block system, which energy of the $LC$ circuit would be analogous to potential energy and which one analogous to kinetic energy?
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Answer

When a charged capacitor $C$ having an initial charge $q_0$ is discharged through an inductance $L,$ the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. The total energy associated with the circuit is constant.
The oscillation of the $LC$ circuit are an electromagnetic analog to the mechanical oscillation of a block$-$spring system.
The total energy of the system remains conserved.
$\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{LI}^2=\text{Constant}=\frac{1}{2}\text{CV}^2_0=\frac{1}{2}\text{LI}^2_0$

Mass spring systerm
v/s
$LC$ circuit
Displacement $(x)$
 
Charge $(q)$
Velocity $(v)$
 
Current $(i)$
Acceleration $(a)$
 
Rate of change of current $\Big(\frac{\text{di}}{\text{dt}}\Big)$
Mass $(m) [$Inertia$]$
 
Inductance $(L) [$Inertia of electricity$]$
Momentum $(p = mv)$
 
Magnetic flux $(\phi=\text{Li})$
Retarding force $\Big(-\text{m}\frac{\text{dv}}{\text{dt}}\Big)$
 
Self induced emf $\Big(-\text{L}\frac{\text{di}}{\text{dt}}\Big)$
 
Equation of free oscillations:
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\omega^2\text{x};$
where $\omega=\sqrt{\frac{\text{K}}{\text{m}}}$
 
Equation of free oscillations:
$\frac{\text{d}^2\text{q}}{\text{dt}^2}=-\Big(\frac{1}{\text{LC}}\Big).\text{q};\text{where }\omega^2=\frac{1}{\text{LC}}$
$\Rightarrow\ \omega=\frac{1}{\sqrt{\text{LC}}}$
Force constant $K$
 
Capacitance $C$
Kinetic energy $=\frac{1}{2}\text{mv}^2$
 
Magnetic energy $=\frac{1}{2}\text{Li}^2$
Elastic potential energy $=\frac{1}{2}\text{Kx}^2$
 
Elecrical potential energy $=\frac{1}{2}\frac{\text{q}^2}{\text{C}}$
If we consider an $L-C$ circuit analogous to a harmonically oscillating spring block system. The electrostatic energy $\frac{1}{2}\text{CV}^2$ is analogous to potential energy and energy associated with moving charges $($current$)$ that is magnetic energy $\Big(\frac{1}{2}\text{LI}^2\Big)$ is analogous to kinetic energy.

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