LCR Circuit

An LCR circuit is an electrical circuit consisting of three components: an inductor (L), a capacitor (C), and a resistor (R), typically connected in series or parallel. These circuits are important for understanding resonance, impedance, and energy oscillations in AC circuits.LCR circuits are foundational in understanding AC behavior, resonance, and the interaction of inductance, capacitance, and resistance.LCR circuit used in Tuning circuits in radios and televisions,Filters in signal processing,Oscillators in communication systems.

• A circuit containing a series combination of an resistance R, a coil of inductance L and a capacitor of capacitance C, connected with a source of alternating e.m.f. of peak value $E_0$.

1.0Circuit Diagram of Series LCR Circuit

2.0Phasor Diagram Of Series LCR Circuit

Let in series LCR circuit applied alternating emf is $E=E_0\sin (\omega t)$. As L, C and R are joined in series,as a result ,the current flowing through the three elements at any given moment has the same amplitude and phase.

However, voltage across each element bears a different phase relationship with the current

3.0Potential difference across L,C and R 

$V_L=I{X}_L{V}_C=I{X}_C{V}_R=IR$

• Now, V_R is in phase with current I but V_L leads I by 90°,while V_C lags behind I by 90°.
• The vector OP represents V_R (which is in phase with I) , the vector OQ represents V_L (which leads I by 90°) and the vector OS represents V_C (which legs behind I by 90°) V_L and V_C are opposite to each other.                                                  

If V_L > V_C (as shown in figure) then their resultant will be (V_L – V_C) which is represented by OT.  Finally, the vector OK represents the resultant of V_R and (V_L – V_C), that is, the resultant of all the three applied e.m.f

$E=\sqrt{V_R^2+(V_L-V_C{)}^2}=\sqrt{I^2{R}^2+(X_L-X_C{)}^2}$

$I=\frac{E}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}$

Phasor diagram also shown that in LCR circuit the applied e.m.f leads the current Iby a phase angle $\varphi$

$\tan \varphi =\frac{X_L-X_C}{R}$

4.0ImpedanceZ in Series LCR Circuit

• The total effective opposition offered by LCR Circuit to alternating current is known as impedance.
• Impedance comprises three parts ResistanceR,Inductive Reactance ($X_L$), Capacitive Reactance ($X_C$), where $X_L$ and $X_C$ are opposite to each other.
• In series LCR Circuit ,the total reactance is taken as $\mp (X_L-X)$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}$

• Unit of Impedance is Ohm

5.0Power Consumed In Series LCR Circuit

• Power dissipated in an a.c circuit is the product of rms value of  voltage and component of current in phase with rms voltage.In series LCR  Circuit ,the phase difference between current and voltage be $\varphi$
• Instantaneous value of  voltage and current in LCR circuit are given by

$V=V_0\sin (\omega t)I=I_0\sin (\omega t+\varphi )$

• Instantaneous Power input to LCR Circuit is given as,

$P_i=VI=V_0{I}_0\sin (\omega t)\sin (\omega t+\varphi )$

$P_i=V_0{I}_0\left({\sin }^2(\omega t)\cos \varphi +\sin (\omega t)\cos (\omega t)\sin \varphi \right)$

$P_i=V_0{I}_0[{\sin }^2\omega t\cos \varphi +\sin \omega t\cos \omega t\sin \varphi ]$

$$\begin{gathered} P_i=V_0{I}_0[{\sin }^2\omega t\cos \varphi +\frac{2\sin \omega t\cos \omega t\sin \varphi }{2}] \\ {P}_i=V_0{I}_0[{\sin }^2\omega t\cos \varphi +\frac{\sin 2\omega t\sin \varphi }{2}] \end{gathered}$$

Average power over a complete cycle of a.c through LCR Circuit is given by,

$P=\frac{{\int }_0^T{P}_{i}dt}{T}$

$P=\frac{1}{T}{\int }_0^T{V}_0{I}_0[{\sin }^2\omega t\cos \varphi +\frac{\sin 2\omega t\sin \varphi }{2}]dt$

$$\begin{gathered} P=\frac{V_0{I}_0}{T}\{{\int }_0^T{\sin }^2\omega t\cos \varphi dt+{\int }_0^T\frac{\sin 2\omega t\sin \varphi }{2}dt\} \\ P=\frac{V_0{I}_0}{T}\{\cos \varphi {\int }_0^T{\sin }^2\omega tdt+\frac{\sin \varphi }{2}{\int }_0^T\sin 2\omega dt\} \end{gathered}$$

${\int }_0^T{\sin }^2\omega tdt={\int }_0^T(\frac{1-\cos 2\omega t}{2})dt=\frac{1}{2}[{\int }_0^{T}dt-{\int }_0^T\cos 2\omega tdt]=\frac{1}{2}[T-0]=\frac{T}{2})$

${\int }_0^T\sin 2\omega dt=0$

$P=\frac{V_0{I}_0}{T}\left\{\cos \varphi \frac{T}{2}+\frac{\sin \varphi }{2}(0)\right\}=\frac{V_0{I}_0}{T}\cdot \frac{T}{2}\cos \varphi =\frac{V_0{I}_0}{2}\cos \varphi$

$P=\frac{V_0}{\sqrt{2}}\cdot \frac{I_0}{\sqrt{2}}\cos \varphi$ 

$P=V_{rms}{I}_{rms}\cos \varphi \text{ where }\cos \varphi \text{ is called the power factor.}$