LCR Series Circuit

A series LCR circuit is a basic but important type of electrical circuit that includes an inductor (L), a capacitor (C), and a resistor (R) all connected in a single loop. It's commonly studied because it shows how these components work together when an alternating current (AC) flows through the circuit. By understanding how the LCR circuit behaves, we can explore key concepts like impedance (resistance to AC), resonance (when the circuit naturally oscillates at a specific frequency), and energy loss. These ideas are crucial in real-world technologies such as tuning radios, filtering signals, and designing electronic systems.

1.0LCR Series Circuit

A series LCR circuit is made by connecting a resistor, an inductor, and a capacitor in one continuous loop. This setup helps us understand how these three components work together when electricity flows back and forth, like in AC (alternating current) systems.

2.0Circuit Diagram Series LCR Circuit

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 of $E_0$.

3.0Phasor Diagram Series LCR Circuit

Let in series an LCR circuit applied alternating emf is $E=E_{0}sin\omega t$. As L, C and R are joined in series, therefore, current at any instant through the three elements has the same amplitude and phase.

4.0Voltage-Current Phase Relationship in LCR Elements

• At any instant t let the current in the circuit be I
• The potential differences across the inductor (L), capacitor (C), and resistor (R) are

$V_L=I{X}_L$  (Inductive Reactance Voltage)

$V_C=I{X}_C$  (Capacitive Reactance Voltage)

$V_R=IR$ (Resistive Voltage)

• The voltage across the resistor $V_R$​ is in phase with the current I
• The voltage across the inductor $V_L$ leads the current I by 90°.
• The voltage across the capacitor $V_C$ lags behind the current I by 90°.

• Vector OP represents $V_R$ (in phase with current).
• Vector OQ represents $V_L$ ​ (leading current by 90°).
• Vector OS represents $V_C$ (lagging current by 90°).
• Since $V_L$ and $V_C$ are opposite in direction, their resultant is $V_L-V_C$, shown as vector OT
• The resultant voltage OK is the vector sum of $V_R$​ and the net reactance voltage $V_L-V_C$, which equals the applied electromotive force (e.m.f) from the AC source.

The resultant of all the three = Applied e.m.f.

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

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

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

The phasor diagram also showed that in LCR circuit the applied e.m.f.

Leads the current I by a phase angle $\varphi$, $Tan\varphi =\frac{X_L-X_C}{R}$

5.0Analysis Series LCR Circuit

1. Circuit diagram

2. Phasor Diagram

a. If $V_L>V_C$  

b. If $V_C>V_L$

c.

$V=\sqrt{V_R^2+{\left(V_L-V_C\right)}^2}$

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

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

3. Impedance Triangle 

4. If $X_L>X_C$ then V leads I, $\varphi$ (Positive) circuit nature Inductive.
5. If $X_C>X_L$ then V lags I, $\varphi$ (Negative) circuit nature Capacitive.
6. In A.C. circuit voltage for L or C may be greater than source voltage or current but it happens only when circuit contains L and C both and on R it is never greater than source voltage or current.
7. In parallel A.C. circuit phase difference between IL and IC is .

6.0Resonance

A circuit is said to be resonant when the natural frequency of circuit is equal to frequency of the applied voltage. For resonance both L and C must be present in the circuit.

Series Resonance

(a) $X_L=X_C$ 

(b) $V_L=V_C$

(C)  f = 0 (V and I in same phase)

(d) $Z_{\min }=R$ (impedance minimum)

(e) $I_{\max }=\frac{V}{R}$   (current maximum)

Resonance Frequency

$X_L=X_C$

${\omega }_{r}L=\frac{1}{{\omega }_{r}C}\Rightarrow {\omega }_r^2=\frac{1}{LC}\Rightarrow {\omega }_r=\frac{1}{\sqrt{LC}}\Rightarrow f_r=\frac{1}{2\pi \sqrt{LC}}$

Variation of Z with f

1. If $f<f_r,then{X}_L<X_C\Rightarrow$Circuit nature is capacitive ,Negative
2. If $f=f_r,then{X}_L=X_C\Rightarrow$ Circuit nature is Resistive ,Zero
3. If $f>f_r,then{X}_L>X_C\Rightarrow$Circuit nature is Inductive ,Positive

Note: Variation of I with f as f increases, Z first decreases then increases.

As f increase, I first increase then decreases

At resonance, the impedance of a series LCR circuit is minimum, allowing it to easily pass the current of its natural frequency. That’s why it’s called an acceptor circuit. This principle is used in radio and TV tuning to select the desired station by matching the circuit’s frequency with the station’s frequency.

Half Power Frequencies-The frequencies at which, power becomes half of its maximum value called half power frequencies

Band width = $f=f_2–f_1$

Quality Factor (Q)

• The Q-factor of an AC circuit basically gives an idea about stored energy & lost energy.$Q=2\pi \frac{maximumenergystoredpercycle}{maximumenergylosspercycle}$
• It represents the sharpness of resonance.
• It is unit less and dimension less quantity.

$Q=\frac{(X_L{)}_r}{R}=\frac{(X_C{)}_r}{R}=\frac{2\pi f_{r}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{f_r}{\Delta f}=\frac{f_r}{\text{Bandwidth}}$

Magnification

• At Resonance $V_{L}or{V}_C=QE$ (where E = supplied voltage)
• So, at Resonance, Magnification factor = Q-factor

Sharpness

Sharpness ∝  Quality factor ∝    Magnification factor

$\text{R decrease}\Rightarrow \text{Q increases}Rightarrow\text{Sharpness increases}$ 

7.0Average Power in Series LCR  Circuit

$V=V_0\sin \omega t\text{and}I=I_0\sin (\omega t-\Phi )$

Instantaneous Power, $P=\left(V_0\sin \omega t\right)\left(I_0\sin (\omega t-\Phi )\right)$

$P=V_0{I}_0\sin \omega t(\sin \omega t\cos \Phi -\sin \Phi \cos \omega t)$

Average Power $⟨P⟩=\frac{1}{T}{\int }_0^T\left(V_0{I}_0{\sin }^2\omega t\cos \Phi -V_0{I}_0\sin \omega t\cos \omega t\sin \Phi \right)dt$

$=V_0{I}_0\left[\frac{1}{T}{\int }_0^T{\sin }^2\omega t\cos \Phi dt-\frac{1}{T}{\int }_0^T\sin \omega t\cos \omega t\sin \Phi dt\right]$

$=V_0{I}_0\left[\frac{1}{2}\cos \Phi -0\times \sin \Phi \right]$

$⟨P⟩=\frac{V_0{I}_0\cos \Phi }{2}=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \Phi$