Question

In a series LCR circuit, L = 10 mH, C = 1 µF and R = 0.4 Ω. i) Write the equation of
motion when the charged capacitor discharges and discuss the nature of the discharge.
ii) Will it be oscillatory or dead beat? iii) How long does the charge oscillations take to
decay to half its initial value.

Answer 1

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.

========================
      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).