# 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.

1
by aarzu

2015-04-14T01:04:27+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.

======
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).

Q(0) = - √(R² + 4L²) ω' Cos δ / [ ω'² - R²/(4L²) ]
Q(t1) = Q(0) / 2 or - Q(0) /2 when ?
=> e^{- R/2L t1 } [ Cos (ω' t1 - δ) ] = 1/2 Cos δ = 0.02489
=> e^{-20 t1 } Cos [ 9999.98 t1 - 1.521 radians ] = 0.02489
=> e^{-20 t1 } Cos [ 9999.98 t1 - 1.521 radians ] = 0.02489
=> 9999.98 t1 - 1.521 radians ≈ 1.5459 rad. as t1 is very small. So exponential term ≈ 1
=> t1 = 0.3067 milliseconds
So it will take 0.3067 milliseconds for the charge to become half its initial value.
CORRECTION: Please delete the last 3 comments. Formula for Q(t) is good.
time period of damped oscillations : T = 2 pi / ω' = 0.628 milliseconds.
================================
the damping envelope is the exponential term:
e^{- R/2L t1 } = 1/2 when R/2L t1 = Ln 2 => t1 = Ln2 / (R/2L)
t1 = 0.0346 sec.
the charge present in the capacitor becomes 1/2 when t = 0.0346 sec or after about 553 cycles.