LCR Circuit - Phasor Diagram, FAQs

What is LCR circuit?

An LCR circuit (also called an RLC circuit) is an electrical circuit that consists of:

• R - Resistor
• L - Inductor
• C-Capacitor

connected in series to an AC voltage source.

The applied alternating emf is:

$
\varepsilon=\varepsilon_0 \sin (\omega t)
$

Since the components are connected in series, the same current flows through R, L, and C.

1. Voltage across Resistor (VR)

$
V_R=I R
$

• In phase with current
• Phasor is along the direction of current

2. Voltage across Inductor (VL)

$
V_L=I X_L=I \omega L
$

• Leads current by $90^{\circ}$
• Phasor is drawn $\mathbf{9 0}^{\boldsymbol{\circ}}$ anticlockwise from current

3. Voltage across Capacitor (VC)

$
V_C=I X_C=I\left(\frac{1}{\omega C}\right)
$

• Lags current by $90^{\circ}$
• Phasor is drawn $\mathbf{9 0}^{\boldsymbol{\circ}}$ clockwise from current

Also read :

• NCERT notes Class 12 Physics Chapter 7 Alternating Current
• NCERT solutions for Class 12 Physics Chapter 7 Alternating Current
• NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

Phasor Diagram of a Series LCR Circuit

• $V_R$ is horizontal
• $V_L$ is drawn upward
• $V_C$ is drawn downward
• $V_L-V_C$ gives the net reactive voltage

The diagonal of the phasor diagram gives the applied voltage $\varepsilon_0$.

Impedance of Series LCR Circuit

Using Pythagoras rule from the phasor diagram:

$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$

where

$
X_L=\omega L, \quad X_C=\frac{1}{\omega C}
$

Current in the circuit becomes:

$
I_0=\frac{\varepsilon_0}{Z}
$

Phase Difference $\mathbf{(} \mathbf{\Phi}$ ) Between Voltage and Current

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

• If $X_L>X_C \rightarrow$ Inductive circuit
• If $X_L<X_C \rightarrow$ Capacitive circuit
• If $X_L=X_C \rightarrow$ Purely resistive circuit ( $\phi=0$ )

Also read -

• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for All Subjects

Resonance in Series LCR Circuit

A series LCR circuit is in resonance when:

$
X_L=X_C
$

This gives:

$
\omega_0=\frac{1}{\sqrt{L C}}
$

Resonant frequency:

$
f_0=\frac{1}{2 \pi \sqrt{L C}}
$

At Resonance:

• $Z=R$ (minimum impedance)
• $I=I_{\text {max }}=\frac{\varepsilon_0}{R}$
• Voltage and current are in phase
• Circuit behaves like a pure resistor
• Power factor $=1$

Application: Radio tuning circuits, filters, wireless communication.

Power Consumed in a Series LCR Circuit

Let the applied alternating voltage be:

$
v=V_0 \sin \omega t
$

The resulting current in the circuit is:

$
i=I_0 \sin (\omega t-\phi)
$

where $\phi$ is the phase difference between voltage and current.

The instantaneous power supplied to the circuit is:

$
\begin{gathered}
p=v \cdot i \\
p=V_0 \sin \omega t \cdot I_0 \sin (\omega t-\phi)
\end{gathered}
$

Using the trigonometric identity:

$
\sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]
$

we get:

$
p=\frac{V_0 I_0}{2}[\cos \phi-\cos (2 \omega t-\phi)]
$

The average value of $\cos (2 \omega t-\phi)$ over one complete cycle is zero.
Hence, the average power consumed in the circuit is:

$
P=\frac{V_0 I_0}{2} \cos \phi
$

$\begin{aligned} V_{r m s} & =\frac{V_0}{\sqrt{2}}, \quad I_{r m s}=\frac{I_0}{\sqrt{2}} \\ P & =V_{r m s} I_{r m s} \cos \phi\end{aligned}$

Power Factor

The quantity $\cos \phi$ is called the power factor of the circuit.
For a series LCR circuit:

$
\cos \phi=\frac{R}{Z}
$

where the impedance $Z$ is given by:

$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$

Thus, the average power can also be written as:

$
P=I_{r m s}^2 R
$

Characteristics of Series Resonance

1. Occurs when $X_L=X_C$.
2. Impedance is minimum (equal to R ).
3. Current is maximum.
4. Voltage across $L$ and $C$ are equal and opposite.
5. Power dissipation is maximum.
6. Power factor $=1$.

NCERT Physics Notes:

• NCERT Notes Class 11th Physics
• NCERT Notes Class 12th Physics
• NCERT Notes For All Subjects

Quality Factor (Q Factor)

The Q-factor shows how sharp the resonance is.

$
Q=\frac{\omega_0 L}{R}=\frac{1}{\omega_0 C R}
$

Also:

$
Q=\frac{f_0}{\Delta f}
$

• High Q → sharper resonance (good selectivity)
• Low Q → broad resonance

Bandwidth of LCR Circuit

$
\Delta f=f_2-f_1=\frac{f_0}{Q}
$

where $f_1$ and $f_2$ are frequencies at which current drops to $I_{\max } / \sqrt{2}$ (half-power frequencies).

Uses of LCR Circuit

• Radio receivers
• Tuned circuits
• Signal processing
• Filtering specific frequencies
• Wireless transmitters
• Sensors and oscillators

Also check-

• NCERT Exemplar Class 11th Physics Solutions
• NCERT Exemplar Class 12th Physics Solutions
• NCERT Exemplar Solutions for All Subjects