Class 12 Physics Alternating Current Series LCR circuit

Series  LCR  circuit

  • LCR circuit consists of inductor having inductance = L, a capacitor having capacitance = C  and  a  resistor having  resistance  = R.
  • As  resistor, capacitor and inductor all are connected in series therefore same amount of current will flow through each of them.
  • Considering source voltage V = Vm sinωt
  • Applying  Kirchhoff’s loop rule to this circuit :-
  • Net  EMF = V (source voltage) + e (self-induced emf) = IR (voltage drop across the resistor) + (q/C) ( voltage drop across the capacitor).
  • =>  Vm sinωt  - L (dI/dt) = IR +(q/C)
  • => Vm sinωt  = IR +(q/C)  + L (dI/dt)
  • By putting I = (dI/dt) :-  Vm sinωt  = R (dq/dt) + (q/C) + L(d2q/dt2)
  • By rearranging , L(d2q/dt2) + R (dq/dt) + (q/C) =  Vm sinωt   (equation(1))

 

 

 Impedance in a LCR circuit

  • Resistance associated with the series LCR circuit  is known as impedance.
  • Impedance is the net resistance offered by the LCR circuit i.e. it includes the resistance offered by the resistor, inductor and the capacitor.
  • It is denoted by Z.
  • SI  unit is  ohm(Ω).
  • The value of Z is given as :- Z = √(R2 + (XC – XL)2)

Impedance diagram:-

  • It is a right angle triangle whose hypotenuse is represented by Z, base is R and the height or perpendicular is (XC – XL)2 .
  • Φ = phase angle  between V(source voltage) and I(current flowing through the circuit).
  • I parallel to VR. Therefore Φ =  angle between the V and VR.
  • Z = √(R2 + (XC – XL)2)  and  tan Φ = (XC – XL)/(R)

 

Case1:- XC > XL

  • (XC – XL) will be positive. Therefore Φ = (+ive).
  • Circuit will be a capacitive circuit  because XC is more.
  • Current(I) will lead the voltage(V) .

Case 2:- XL > Xc

  • (XC – XL) will be negative. Therefore Φ =(-ive).
  • Circuit will be an  inductive  circuit  because  XL is more.
  • Voltage(V) will lead  the current(I).

 AC  voltage applied  to  a Series  LCR  circuit

  • In order to solve the given equation    L(d2q/dt2) + R (dq/dt) + (q/C) =  Vm sinωt   (equation(1)), we have to assume a solution for this:- q = qm sin (ωt + q )
  • By Differentiating, (dq/dt) = qm ω cos(ωt + q)  , and
  •  (d2q/dt2) = - qm ω2 sin(ωt + q) ;
  • Substituting above values  in (equation(1)) and simplifying we get,
  •  => qm ω [R cos (ωt + q) + (XC - XL) sin(ωt + q)] = Vm sinωt  
  • Divide and multiply by Z  throughout the equation:-
    • Where Z = Impedance
  • => (qm ω Z) =  [(R/Z) cos (ωt + q) + (1/Z)( XC - XL) sin (ωt + q) ] = Vm sinωt  
    • Therefore expression becomes,(using (R/Z) = cos f and  (XC - XL)/(Z) = sin f)
  • (qm ω Z)  [cos f cos (ωt + q) + sin f sin (ωt + q)] = Vm sinωt  
  • (qm ω Z) [cos ( - f + ωt + q) = Vm sinωt  
  • Therefore, Vm  = Z  Im (replacing (qm ω) = Im) ,
  • After calculating and simplifying, we get
  • I = Im sin(ωt + f) . This is the expression for current in series LCR circuit.

Points to be noted:-

  1. Voltage and current are in/out of phase  depends on f.
  2. Current Amplitude is given by: Im = qm ω.
  • This is because V = Vm sinωt and current I = Im sin(ωt + f).
  • If f =0 then voltage and current are  in phase with each other.
  • If f = (∏/2) then voltage and current are out of phase with each other.
  1. Equation  for  series  LCR  circuit  resembles  that of a forced, damped  oscillator.

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