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)
- 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:-
- => (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:-
- Voltage and current are in/out of phase depends on f.
- 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.
- Equation for series LCR circuit resembles that of a forced, damped oscillator.