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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