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 = V
_{m}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).
- => V
_{m}sinωt - L (dI/dt) = IR +(q/C) - => V
_{m}sinωt = IR +(q/C) + L (dI/dt) - By putting I = (dI/dt) :- V
_{m}sinωt = R (dq/dt) + (q/C) + L(d^{2}q/dt^{2}) - By rearranging ,
**L(d**(equation(1))^{2}q/dt^{2}) + R (dq/dt) + (q/C) = V_{m}sinωt

__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 = √(R
^{2}+ (X_{C}– X_{L})^{2})

__Impedance diagram:-__

- It is a right angle triangle whose hypotenuse is represented by Z, base is R and the height or perpendicular is (X
_{C}– X_{L})^{2}. - Φ = phase angle between V(source voltage) and I(current flowing through the circuit).
- I parallel to V
_{R}. Therefore Φ = angle between the V and V_{R}. **Z = √(R**and tan Φ = (X^{2}+ (X_{C}– X_{L})^{2})_{C}– X_{L})/(R)

**Case1:**- X_{C} > X_{L}

- (X
_{C}– X_{L}) will be positive. Therefore Φ = (+ive). - Circuit will be a capacitive circuit because X
_{C}is more. - Current(I) will lead the voltage(V) .

**Case 2**:- X_{L} > X_{c}

- (X
_{C}– X_{L}) will be negative. Therefore Φ =(-ive). - Circuit will be an inductive circuit because X
_{L}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(d
^{2}q/dt^{2}) + R (dq/dt) + (q/C) = V_{m}sinωt (equation(1)), we have to assume a solution for this:- q = q_{m}sin (ωt + q ) - By Differentiating, (dq/dt) = q
_{m }ω cos(ωt + q) , and - (d
^{2}q/dt^{2}) = - q_{m }ω^{2}sin(ωt + q) ; - Substituting above values in (equation(1)) and simplifying we get,
- => q
_{m}ω [R cos (ωt + q) + (X_{C}- X_{L}) sin(ωt + q)] = V_{m}sinωt - Divide and multiply by Z throughout the equation:-
- Where Z =
__Impedance__ - => (q
_{m}ω Z) = [(R/Z) cos (ωt + q) + (1/Z)( X_{C}- X_{L}) sin (ωt + q) ] = V_{m}sinωt - Therefore expression becomes,(using (R/Z) = cos f and (X
_{C}- X_{L})/(Z) = sin f)

- (q
_{m}ω Z) [cos f cos (ωt + q) + sin f sin (ωt + q)] = V_{m}sinωt - (q
_{m}ω Z) [cos ( - f + ωt + q) = V_{m}sinωt - Therefore,
**V**(replacing_{m}= Z I_{m }**(q**) ,_{m}ω) = I_{m} - After calculating and simplifying, we get
**I = I**_{m}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: I
_{m}= q_{m}ω.

- This is because V = V
_{m}sinωt and current I = I_{m}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.

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