Suppose the capacitor is charged to Q₀ at t= 0.

We have L = 10 mH , C = 1 μF , R = 0.4 Ω

When at t =0, the circuit with L , C, and R is closed, then the capacitor starts to discharge. The current starts to flow in inductor and stores energy in it. Some energy starts dissipating through R.

sum of potential diff, is 0 in the circuit. let I be the current in the circuit.

L d I / dt + Q/C + R I = 0 differentiate wrt t, and divide by L

* d² I /d t² + I / LC + R/L d I / dt = 0* ---- (1)

let b = R = 0.4Ω , ω₀² = 1/LC = k/m, so k =1/C = 10⁶ F⁻¹ ,and m = L = 10 mH

So we have ω₀ = 10⁴ rad/sec , f = 10⁴/2π Hz

then * d² I/ dt² + b/m d I / dt + ω₀² I = 0 * this is the equation.

This looks exactly like the equation for a damped oscillator, with a natural frequency = ω₀.

we know the solution : I (t) = I₀ e^{- R/2L t } Sin (ω' t + Ф')

where ω' = √[ ω₀² - (R/2L)² ] = √ [ 10⁸ - 20² ] = 9999.98 rad/sec

* as R/2L is < < ω₀, the circuit will oscillate as a slow damped circuit.*

(condition : R < 2 √[L/C] same as above)

So * I (t) = I₀ e^{- R/2L t } Sin (ω' t + Ф')*

I (t = 0) = I₀ Sin Ф'

where I₀ =

Cos Ф' = b/(2m ω₀) = R / (2 L ω₀) = 20/10000 = 0.002

Ф' = 1.568 radians or 89.88 deg.

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Quality factor of the circuit is = Qf = √[L/C] / R = ω₀ / (b/m) = ω₀ / (R/L)

= 250 approximately

Qf gives an indication of number of - minimum - cycles that oscillations will take place, before the amplitude becomes low.

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d Q(t) / dt = * I (t) = I₀ e^{- R/2L t } Sin (ω' t + Ф')*

we use the integration by parts: u = e^{- R/2L t } v ' = Sin (ω' t + Ф')

v = - 1/ω' Cos (ω' t + Ф')

Q(t) = - 1/ω' e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω') integral e^{- R/2L t } Cos (ω' t + Ф')

= - 1/ω' e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω') [ e^{- R/2L t } Sin (ω' t + Ф')

- R/(2Lω') integral e^{- R/2L t } Sin (ω' t + Ф') ]

= - 1/ω' e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω') e^{- R/2L t } Sin (ω' t + Ф')

+ R²/(4L²ω'²) Q(t)

Q(t) [ 1 - R²/(4L²ω'²) ] = - 1/ω' e^{- R/2L t } Cos(ω' t+Ф') - R/(2Lω') e^{- R/2L t } Sin (ω' t + Ф')

Q(t) = - 1/ω' e^{- R/2L t } [ Cos(ω' t+Ф') + R/(2L) Sin (ω' t + Ф') ] / [ 1 - R²/(4L²ω'²) ]

Q(0) = - 1/ω' [ Cos Ф' + R/2L Sin Ф' ] / [ 1 - R²/(4L²ω'²) ]

we want time t such that the amplitude is half of Q(0).