Series LCR Circuits and Resonance

Ac Voltage Applied To A Series Lcr Circuit (Impedance $ Z = \sqrt{R^2 + (X_L - X_C)^2} $)

A series LCR circuit is an electric circuit containing a resistor ($R$), an inductor ($L$), and a capacitor ($C$) connected in series across an alternating voltage source. This is a fundamental circuit in AC analysis, as it exhibits the phenomena of resonance and impedance, which are crucial for understanding many electronic circuits like filters and oscillators.

---

Circuit Setup

Consider a circuit where a resistor $R$, an inductor $L$, and a capacitor $C$ are connected in series to an AC voltage source. The instantaneous voltage supplied by the source is $v(t) = V_0 \sin(\omega t)$.

A series LCR circuit connected to an AC voltage source.

In a series circuit, the instantaneous current $i(t)$ flowing through each component (R, L, and C) is the same at any instant. Let this current be $i(t) = I_0 \sin(\omega t + \phi)$, where $I_0$ is the peak current and $\phi$ is the phase difference between the voltage and current relative to the source voltage phase. We need to determine $I_0$ and $\phi$.

The instantaneous voltage across each component will have a specific phase relationship with this current:

• Voltage across resistor: $v_R(t)$. In a resistor, voltage is in phase with current. $v_R(t) = i(t)R$. Peak voltage $V_{R0} = I_0 R$.
• Voltage across inductor: $v_L(t)$. In a pure inductor, voltage leads current by $90^\circ$. $v_L(t) = I_0 X_L \sin(\omega t + \phi + \pi/2)$. Peak voltage $V_{L0} = I_0 X_L$.
• Voltage across capacitor: $v_C(t)$. In a pure capacitor, voltage lags current by $90^\circ$. $v_C(t) = I_0 X_C \sin(\omega t + \phi - \pi/2)$. Peak voltage $V_{C0} = I_0 X_C$.

According to Kirchhoff's Loop Rule (KVL), the instantaneous voltage across the source is equal to the sum of the instantaneous voltages across the individual components:

$ v(t) = v_R(t) + v_L(t) + v_C(t) $

Adding these sinusoidal functions with phase differences directly in terms of time functions is complex. This is where the phasor approach becomes very helpful.

---

Phasor-Diagram Solution

We can represent the instantaneous current and the voltages across R, L, and C by phasors. Since the current is the same through all components in series, it's convenient to choose the current phasor as the reference. Let the current phasor $\vec{I}$ have magnitude $I_0$ and be along the horizontal axis (representing phase $\omega t + \phi = \omega t_0$ at a reference time $t_0$).

Phasor diagram for a series LCR circuit. Current phasor $\vec{I}$ is the reference. $\vec{V}_R$ is in phase with $\vec{I}$, $\vec{V}_L$ leads $\vec{I}$ by 90°, $\vec{V}_C$ lags $\vec{I}$ by 90°.

The voltage phasors for each component are:

• Resistor Voltage Phasor ($\vec{V}_R$): Magnitude $V_{R0} = I_0 R$. Since $v_R$ is in phase with $i$, $\vec{V}_R$ is in the same direction as $\vec{I}$ (along the horizontal axis).
• Inductor Voltage Phasor ($\vec{V}_L$): Magnitude $V_{L0} = I_0 X_L$. Since $v_L$ leads $i$ by $90^\circ$, $\vec{V}_L$ is rotated $90^\circ$ counter-clockwise relative to $\vec{I}$ (along the positive vertical axis).
• Capacitor Voltage Phasor ($\vec{V}_C$): Magnitude $V_{C0} = I_0 X_C$. Since $v_C$ lags $i$ by $90^\circ$, $\vec{V}_C$ is rotated $90^\circ$ clockwise relative to $\vec{I}$ (along the negative vertical axis).

According to KVL, the phasor representing the source voltage $\vec{V}$ is the vector sum of the individual voltage phasors:

$ \vec{V} = \vec{V}_R + \vec{V}_L + \vec{V}_C $

In the phasor diagram, $\vec{V}_L$ and $\vec{V}_C$ are opposite to each other (along the vertical axis). Their vector sum is a phasor along the vertical axis with magnitude $|V_{L0} - V_{C0}|$. The net vertical phasor is $(\vec{V}_L + \vec{V}_C)$.

The source voltage phasor $\vec{V}$ is the vector sum of the horizontal phasor $\vec{V}_R$ and the net vertical phasor $(\vec{V}_L + \vec{V}_C)$. These two vectors are perpendicular.

The magnitude of the source voltage phasor, which is the peak source voltage $V_0$, is found using the Pythagorean theorem:

$ V_0^2 = V_{R0}^2 + (V_{L0} - V_{C0})^2 $

Substitute the peak voltage expressions in terms of current:

$ V_0^2 = (I_0 R)^2 + (I_0 X_L - I_0 X_C)^2 $

$ V_0^2 = I_0^2 R^2 + I_0^2 (X_L - X_C)^2 $

$ V_0^2 = I_0^2 [R^2 + (X_L - X_C)^2] $

Taking the square root of both sides and solving for $I_0$:

$ I_0 = \frac{V_0}{\sqrt{R^2 + (X_L - X_C)^2}} $

---

Impedance ($Z$)

Comparing the peak voltage-current relationship ($V_0 = I_0 \sqrt{R^2 + (X_L - X_C)^2}$) with Ohm's Law ($V_0 = I_0 R$), the term $\sqrt{R^2 + (X_L - X_C)^2}$ represents the overall opposition offered by the series LCR circuit to the flow of alternating current. This quantity is called the impedance ($Z$) of the circuit.

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

Impedance is the AC equivalent of resistance in DC circuits. It is a measure of the total opposition to current flow, considering resistance, inductive reactance, and capacitive reactance. The unit of impedance is the ohm ($\Omega$).

The relationship between peak voltage and current (or RMS values) in a series LCR circuit is given by:

$ V_0 = I_0 Z \quad \text{or} \quad V_{rms} = I_{rms} Z $

---

Phase Difference ($\phi$)

From the phasor diagram, the angle $\phi$ between the source voltage phasor $\vec{V}$ and the current phasor $\vec{I}$ represents the phase difference between the applied voltage and the total current flowing in the circuit.

From the right triangle formed by $\vec{V}_R$, $(\vec{V}_L + \vec{V}_C)$, and $\vec{V}$:

$ \tan\phi = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{V_{L0} - V_{C0}}{V_{R0}} $

Substitute the peak voltage expressions in terms of current:

$ \tan\phi = \frac{I_0 X_L - I_0 X_C}{I_0 R} = \frac{I_0 (X_L - X_C)}{I_0 R} $

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

The phase difference $\phi$ is given by $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$.

• If $X_L > X_C$, then $\tan\phi$ is positive, and $\phi$ is positive. This means the voltage leads the current ($\phi > 0$), and the circuit is predominantly inductive.
• If $X_C > X_L$, then $\tan\phi$ is negative, and $\phi$ is negative. This means the voltage lags the current ($\phi < 0$), and the circuit is predominantly capacitive (current leads voltage).
• If $X_L = X_C$, then $\tan\phi = 0$, and $\phi = 0$. This means the voltage and current are in phase, and the circuit is purely resistive. This condition is called resonance.

---

Analytical Solution

The analysis using complex numbers or differential equations is another way to solve AC circuits. Let the applied voltage be $v(t) = V_0 \sin(\omega t)$. We are looking for the steady-state current $i(t) = I_0 \sin(\omega t + \phi)$.

Applying Kirchhoff's Loop Rule:

$ v_R(t) + v_L(t) + v_C(t) = v(t) $

Substitute the voltage-current relations:

$ iR + L \frac{di}{dt} + \frac{1}{C} \int i \, dt = V_0 \sin(\omega t) $

This is a second-order differential equation. Assuming a solution of the form $i(t) = I_0 \sin(\omega t + \phi)$ and substituting it into the equation leads to the same expressions for $I_0$ and $\phi$ as obtained using the phasor method.

Using impedance concepts with complex numbers provides a more algebraic way to solve this. Representing $v(t) = V_0 e^{j\omega t}$ (real part is $V_0 \cos\omega t$, or using $jV_0 e^{j\omega t}$ for sine convention) and resistances/reactances as complex impedances $Z_R = R$, $Z_L = j X_L = j\omega L$, $Z_C = \frac{1}{j \omega C} = -j \frac{1}{\omega C} = -j X_C$.

For series impedance, the total impedance $Z$ is the sum of individual impedances:

$ Z = Z_R + Z_L + Z_C = R + j X_L - j X_C = R + j (X_L - X_C) $

The magnitude of the impedance is $|Z| = \sqrt{R^2 + (X_L - X_C)^2}$, which is the same impedance $Z$ derived from the phasor diagram.

The phase angle $\phi$ is the argument of the complex impedance:

$ \tan\phi = \frac{\text{Imaginary Part}}{\text{Real Part}} = \frac{X_L - X_C}{R} $

The peak current is $I_0 = V_0 / |Z|$. The phase of the current relative to the voltage is $-\phi$ (if voltage phase is 0 and current phase is $\phi$, then $v$ leads $i$ by $\phi$; if current phase is 0 and voltage phase is $\phi$, then $v$ leads $i$ by $\phi$). In our initial setup, $v(t) = V_0 \sin(\omega t)$, phase is 0. $i(t) = I_0 \sin(\omega t + \phi)$, phase is $\phi$. So current leads voltage if $\phi>0$, lags if $\phi<0$. The derived phase $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$ is the phase difference $\phi_i - \phi_v = \phi - 0 = \phi$. If $X_L > X_C$, $\phi > 0$, current leads voltage. If $X_C > X_L$, $\phi < 0$, current lags voltage. This is the opposite of what was stated earlier.

Let's check conventions. If $v(t) = V_0 \sin(\omega t)$ and $i(t) = I_0 \sin(\omega t + \phi)$, then current leads voltage by $\phi$. Phasor for V at angle 0, phasor for I at angle $\phi$. Angle of I relative to V is $\phi$. In phasor diagram where I is reference (angle 0), V is at angle $\phi$. So $\vec{V}$ is at angle $\phi$ relative to $\vec{I}$. $\tan\phi = \frac{V_{L0}-V_{C0}}{V_{R0}} = \frac{X_L-X_C}{R}$. So $\phi$ is the angle by which voltage leads current. If $\phi$ is positive, voltage leads current. If $\phi$ is negative, voltage lags current (or current leads voltage). The convention is that $\phi$ is the phase of voltage relative to current, or phase of current relative to voltage, depending on the definition. The usual convention is that $\phi$ represents the phase angle by which the voltage across the entire circuit leads the current in the circuit. So if $\phi$ is positive, voltage leads current.

Let's re-verify the phase relationships: Resistor: $v_R = iR$. $v_R$ and $i$ are in phase. Inductor: $v_L = L di/dt$. If $i = I_0 \sin(\omega t)$, $v_L = L I_0 \omega \cos(\omega t) = I_0 X_L \sin(\omega t + \pi/2)$. Voltage leads current by $\pi/2$. Capacitor: $i = C dv_C/dt$. If $v_C = V_{C0} \sin(\omega t)$, $i = C V_{C0} \omega \cos(\omega t) = \frac{V_{C0}}{X_C} \sin(\omega t + \pi/2)$. Current leads voltage by $\pi/2$. Voltage lags current by $\pi/2$.

In the series LCR circuit, if $i(t) = I_0 \sin(\omega t + \phi_i)$, then $v_R$ is in phase, $v_L$ leads by $\pi/2$, $v_C$ lags by $\pi/2$. The sum of these voltages equals $v(t) = V_0 \sin(\omega t)$.

If we take current as reference, $i(t) = I_0 \sin(\omega t)$. Then $v_R(t) = I_0 R \sin(\omega t)$, $v_L(t) = I_0 X_L \sin(\omega t + \pi/2)$, $v_C(t) = I_0 X_C \sin(\omega t - \pi/2)$. The source voltage $v(t) = v_R + v_L + v_C$. This sum is $v(t) = V_0 \sin(\omega t + \phi)$, where $\phi$ is the phase of $v$ relative to $i$. The phasor sum gives $V_0^2 = (I_0R)^2 + (I_0X_L - I_0X_C)^2$ and $\tan\phi = \frac{I_0X_L - I_0X_C}{I_0R} = \frac{X_L - X_C}{R}$. So $\phi$ is the angle by which the applied voltage leads the current.

If $X_L > X_C$, $\phi > 0$, voltage leads current. If $X_C > X_L$, $\phi < 0$, voltage lags current (current leads voltage). If $X_L = X_C$, $\phi = 0$, voltage and current are in phase.

Okay, the previous statement was correct, I got confused by my own example notation. The derived $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$ is indeed the phase angle by which the voltage leads the current.

---

Resonance ($ \omega_0 = 1/\sqrt{LC} $)

In a series LCR circuit, the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$. The current amplitude is $I_0 = V_0/Z = \frac{V_0}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$.

For a given source voltage amplitude $V_0$ and circuit components $R, L, C$, the current amplitude $I_0$ depends on the angular frequency $\omega$ of the source. The current amplitude will be maximum when the impedance $Z$ is minimum.

The impedance $Z$ is minimum when the term $(X_L - X_C)^2$ is minimum. Since it's a square, the minimum value is zero, which occurs when $X_L - X_C = 0$, i.e., $X_L = X_C$.

$ X_L = X_C $

$ \omega L = \frac{1}{\omega C} $

Rearranging for $\omega$:

$ \omega^2 = \frac{1}{LC} $

$ \omega = \frac{1}{\sqrt{LC}} $

This specific angular frequency $\omega_0 = \frac{1}{\sqrt{LC}}$ is called the resonant angular frequency of the series LCR circuit. When the frequency of the applied AC voltage is equal to the resonant frequency of the circuit, the circuit is said to be in resonance.

Resonant frequency $f_0 = \frac{\omega_0}{2\pi} = \frac{1}{2\pi\sqrt{LC}}$

---

Properties of Series Resonance

At resonance ($\omega = \omega_0$):

• $X_L = X_C$.
• Impedance $Z$ is minimum: $Z_{min} = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0^2} = R$. The impedance is purely resistive and equal to the circuit resistance.
• Current amplitude $I_0$ is maximum: $I_{0, max} = V_0/Z_{min} = V_0/R$. This is the largest current possible in the circuit for the given voltage amplitude.
• The phase difference $\phi$ between voltage and current is zero: $\tan\phi = \frac{X_L - X_C}{R} = \frac{0}{R} = 0$. So $\phi = 0$. Voltage and current are in phase.
• The voltage across the inductor and capacitor can be much larger than the applied voltage $V_0$. $V_{L0} = I_{0, max} X_L = (V_0/R) X_L$ and $V_{C0} = I_{0, max} X_C = (V_0/R) X_C$. Since $X_L = X_C$ at resonance, $V_{L0} = V_{C0}$. If $X_L > R$, then $V_{L0}$ and $V_{C0}$ can be greater than $V_0$. This phenomenon is called voltage magnification or voltage resonance.
• Average power factor ($\cos\phi$) is maximum: $\cos 0^\circ = 1$.
• Average power absorbed by the circuit is maximum: $P_{avg} = V_{rms} I_{rms} \cos\phi = V_{rms} I_{rms} (1)$.

A graph of current amplitude ($I_0$) versus frequency ($\omega$) for a series LCR circuit shows a sharp peak at the resonant frequency $\omega_0$.

Resonance curve showing maximum current at resonant frequency $\omega_0$.

---

Sharpness Of Resonance (Q Factor)

The resonance in a series LCR circuit can be sharp or flat. The sharpness of resonance refers to how quickly the current amplitude falls off from its maximum value at resonance as the frequency deviates from the resonant frequency. A sharper resonance means the circuit is very selective in terms of frequency; it passes currents primarily at the resonant frequency and suppresses currents at other frequencies. This property is crucial in tuning radio receivers.

The sharpness of resonance is quantitatively measured by the Quality factor (Q factor) of the circuit.

There are several equivalent definitions for the Q factor of a series LCR circuit:

1. Ratio of resonant frequency to bandwidth: $ Q = \frac{\omega_0}{\Delta\omega} $, where $\Delta\omega$ is the bandwidth (the range of frequencies where the power is at least half of the maximum power, or current is at least $1/\sqrt{2}$ times the maximum current). A higher Q factor means a smaller bandwidth and hence a sharper resonance peak.
2. Ratio of voltage across L or C to voltage across R at resonance:

   $ Q = \frac{V_{L0}}{V_{R0}} = \frac{I_{0, max} X_L}{I_{0, max} R} = \frac{X_L}{R} = \frac{\omega_0 L}{R} $

   or

   $ Q = \frac{V_{C0}}{V_{R0}} = \frac{I_{0, max} X_C}{I_{0, max} R} = \frac{X_C}{R} = \frac{1}{\omega_0 C R} $

   Since $X_L = X_C$ at resonance, $\frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}$.
3. In terms of L, C, and R: Substitute $\omega_0 = 1/\sqrt{LC}$ into either formula from point 2.

   $ Q = \frac{\omega_0 L}{R} = \frac{(1/\sqrt{LC}) L}{R} = \frac{L}{R\sqrt{LC}} = \frac{1}{R} \sqrt{\frac{L}{C}} $

   or

   $ Q = \frac{1}{\omega_0 C R} = \frac{1}{(1/\sqrt{LC}) C R} = \frac{\sqrt{LC}}{C R} = \frac{1}{R} \sqrt{\frac{L}{C}} $

   So, $ Q = \frac{1}{R} \sqrt{\frac{L}{C}} $

A higher Q factor is achieved when the resistance $R$ is low, the inductance $L$ is high, and the capacitance $C$ is low. A low resistance minimises energy dissipation and allows for larger current fluctuations near resonance.

A circuit with a high Q factor has a sharp resonance curve, making it very selective to a narrow range of frequencies around the resonant frequency. A circuit with a low Q factor has a broad resonance curve, meaning it responds to a wider range of frequencies.

---

Answer: