Question

Show that the total energy is conserved during LC oscillations.

Answer 1

Conservation of energy in LC oscillations: During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor. Although, these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.

Total energy,

U = U_E + U_B = `"q"^2/"2C" + 1/2 "Li"^2`

Let us consider 3 different stages of LC oscillations and calculate the total energy of the system.

Case I:

When the charge in the capacitor, q = Q_m and the current through the inductor, i = 0, the total energy is given by

U = `"Q"_"m"^2/"2C" + 0 = "Q"_"m"^2/"2C"`

The total energy is wholly electrical.

Case II:

When charge = 0; current = I_m, the total energy is

U = `0 + 1/2 "LI"_"m"^2 = 1/2 "LI"_"m"^2 = "L"/2 xx (("Q"_"m"^2)/"LC")`    .....since `"I"_"m" = "Q"_"m" omega = "Q"_"m"/sqrt"LC"`

`= "Q"_"m"^2/(2"C")`

The total energy is wholly electrical.

Case III:

When charge = q; current = i, the total energy is

U = `"q"^2/"2C" + 1/2 "Li"^2`

Since q = Q_m cos ωt, i = `"bq"/"dt"` = Q_mω sin ωt. The negative sign in current indicates that the charge in the capacitor in the capacitor decreases with time.

U = `("Q"_"m"^2 cos^2 omega"t")/("2C") + ("L"omega^2"Q"_"m"^2 sin^2 omega"t")/2`

`= ("Q"_"m"^2 cos^2 omega"t")/"2C" + ("L"^2"Q"_"m"^2 sin^2 omega"t")/"2LC"`    .....since `omega^2 = 1/"LC"`

U = `"Q"_"m"^2/"2C" (cos^2 omega"t" + sin^2 omega"t") = "Q"_"m"^2/"2C"`

From above three cases, it is clear that the total energy of the system remains constant.