AC Voltage Applied to a Series LCR Circuit

An electric circuit consisting of a resistor, a constant voltage source, and a key-current is developed when the key is plugged in. The current, flowing in any circuit, has a unique direction, i.e. from the battery’s negative terminal to its positive terminal. If the magnitude of resistance and voltage remains the same, then the current in the circuit will remain constant by Ohm’s law \(\left( {V = IR} \right)\)
But if instead of a battery (constant voltage source), an alternating voltage source is applied across the resistor. Therefore, the current direction will change alternatively across the resistor, and this current can be termed as Alternating current. The source of the alternating voltage can be an AC generator or an AC dynamo.
We normally come across circuits in which, along with resistor, we have other elements like capacitor and inductor too. Several combinations are possible, and each combination serves a different purpose. In this topic, we will learn about AC Voltage applied to a Series LCR circuit, i.e., a circuit consisting of a resistor, capacitor, and an inductor- all connected in series.

What is a Series LCR Circuit ?

An LCR circuit is an electrical circuit consisting of an inductor (L), a capacitor (C), and a resistor (R). It is also known as a tuned or resonant circuit. The combination of these elements, when connected in series, is called a series LCR circuit. Thus, the same amount of current will flow across the resistor, capacitor, and inductor. Here we will study what happens when AC voltage is applied across a series LCR circuit.

The above circuit represents a basic LCR series circuit where a voltage \({V_S}\) is applied across the LCR series circuit. To determine the circuit expressions for the series LCR circuit, we first need to understand what impedance is?
Impedance is the measure of opposition offered by a circuit to the flow of current. It is the effective resistance to the flow of alternating current in an electric circuit consisting of various electric components. It arises from the combined effect of ohmic resistance, capacitive and inductive reactance. If \(R\) represents resistance, \({X_L}\) represents inductive reactance and \({X_C}\) represents capacitive reactance, then the impedance \(Z\) can be given as:
\(Z = \sqrt {{R^2} + \left( {X_C}\,- {X_L} \right)^2} \)
The SI unit of impedance is denoted by \({\rm{\Omega }}\) (Ohm).

Derivation for AC Voltage applied across Series LCR Circuit

To derive the circuit equations for a series LCR circuit, consider the circuit diagram given below:

The electrical circuit consists of an inductor (L), capacitor (C), and resistor (R) connected in series, and an AC voltage supply is connected in the circuit. The voltage source supplies the alternating voltage \(V,\) where,
\(V = {V_m}\sin \left( {\omega t} \right)\)
Where,
\({V_m}\,:\) is the amplitude of the applied voltage
\(\omega \,:\) is the frequency of the applied voltage

If \(q\) be the charge on the capacitor and \(i\) be the current flowing in the circuit at any time \(t,\) then by Kirchhoff’s loop law, the voltage equation of the circuit can be given as:

Net EMF across the circuit: V (source voltage) = Voltage drop across resistor + Voltage drop across capacitor + Self-induced Faraday’s emf in the inductor
\(V = L\left( {\frac{{di}}{{dt}}} \right) + iR + \frac{q}{C}\)
Here, \(L\) is the self-inductance of the inductor. Substituting the expression for alternating voltage,
\({V_m}\sin \left( {\omega t} \right) = L\left( {\frac{{di}}{{dt}}} \right) + iR + \frac{q}{C}\,\,\,\,\,….\left( 1 \right)\)

To determine the instantaneous current \(i\) or its corresponding phase to the applied alternating voltage \(V,\) let us apply the analytical method.

We know that current is equal to the rate of flow of electric charge per unit time, i.e.,

\(i = \frac{{dq}}{{dt}}\)

Differentiating both sides with respect to time, we get:

\(\frac{{di}}{{dt}} = \frac{{{d^2}q}}{{d{t^2}}}\)

Substituting the above value into equation (1), we get the voltage equation in terms of \(q\) as:

\({V_m}\sin \left( {\omega t} \right) = L\left( {\frac{{{d^2}q}}{{d{t^2}}}} \right) + \left( {\frac{{dq}}{{dt}}} \right)R + \frac{q}{C}\,\,\,\,…..\left( 2 \right)\)

This equation is similar to the equation of a forced or damped harmonic oscillator. Let us assume a solution for the above equation,

\(q = {q_m}\sin\left( {\omega t + \theta } \right)\)

Differentiating both sides with respect to time,

\(\frac{{dq}}{{dt}} = {q_m}\,\omega \,\cos\left( {\omega t + \theta } \right)\)

\(\frac{{{d^2}q}}{{d{t^2}}} =  – {q_m}{\omega ^2}\sin\left( {\omega t + \theta } \right)\)

Substituting these values in equation (2),

\({V_m}\sin \left( {\omega t} \right) = {q_m}\omega \left[ {R{\rm{cos}}\left( {\omega t + \theta } \right) + \left( {{X_C} – {X_L}} \right){\rm{sin}}\;\left( {\omega t + \theta } \right)} \right]\,\,\,\,….\left( 3 \right)\)

Here,

Capacitive reactance: \({X_C} = \frac{1}{{\omega C}}\)

Inductive reactance: \({X_L} = \omega L\)

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

Substituting the above values in equation (3), so we get:

\({V_m}\sin \left( {\omega t} \right) = {q_m}\omega Z\left[ {\frac{R}{Z}\cos\left( {\omega t + \theta } \right) + \frac{{\left( {{X_C} – {X_L}} \right)}}{Z}\sin \left( {\omega t + \theta } \right)} \right]\,\,\,…\left( 4 \right)\)

Consider,

\(\frac{R}{Z} ={\rm{cos}}\,\emptyset \)

\(\frac{{\left( {{X_C} – {X_L}} \right)}}{Z} = {\rm{sin}}\,\emptyset \)

Dividing the two equations:

\(\frac{{\left( {{X_C} – {X_L}} \right)}}{R} = {\rm{tan}}\,\emptyset \)

\(\emptyset  = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{\left( {{X_C} – {X_L}} \right)}}{R}} \right)\)

Substituting the above values in equation (4):

\({V_m}\sin \left( {\omega t} \right) = {q_m}\omega Z\left[ {{\rm{cos}}\left( {\omega t + \theta  – \emptyset } \right)} \right]\)

Comparing the LHS and RHS of this equation, we get

\({V_m} = {q_m}\omega Z = {i_m}Z\)

The current in the LCR circuit, \(i = \frac{{dq}}{{dt}}\)

Or, \(i = {q_m}\omega \,{\rm{cos}}\left( {\omega t + \theta } \right)\)

\(i = {i_m}{\rm{cos}}\left( {\omega t + \theta } \right)\)

[where, \({q_m}\omega  = {i_m}\)]

Since, \(\theta  – \emptyset  =  – \frac{\pi }{2}\)

\(\theta  =  – \frac{\pi }{2} + \emptyset \)

We get, \(i = {i_m}{\rm{cos}}\left( {\omega t – \frac{\pi }{2} + \emptyset } \right)\)

\(i = {i_m}{\rm{sin}}\left( {\omega t + \emptyset } \right)\)

Here, \({i_m} = \frac{{{V_m}}}{Z} = \frac{{{V_m}}}{{\sqrt {{R^2} + {{\left( {{X_C} – {X_L}} \right)}^2}} }}\;\) and \(\emptyset = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{{X_C} – {X_L}}}{R}} \right)\)

Thus, for \(\theta = 0^\circ \) : The applied voltage and instantaneous current are in phase.

For \(\theta = 90^\circ \) : The applied voltage and instantaneous current are out-of-phase.

Resonance of LCR Circuit

A given circuit is in resonance if its output becomes maximum at a specific frequency. The resonance phenomenon is associated with the systems that present the tendency to oscillate at a specific frequency called the given system’s natural frequency. If an energy source drives such a system at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large.

For the given series LCR circuit, we found that the amplitude of voltage, frequency, and current are related to each other as:

\({i_m} = \frac{{{V_m}}}{Z} = \frac{{{V_m}}}{{\sqrt {{R^2} + {{\left( {{X_C} – {X_L}} \right)}^2}} }}\)

Where, \({X_C} = \frac{1}{{\omega C}}\) and \({X_L} = \omega L\)

\({i_m} = \frac{{{V_m}}}{Z} = \frac{{{V_m}}}{{\sqrt {{R^2} + {{\left( {\frac{1}{{\omega C}}\,-\,\omega L} \right)}^2}} }}\)

The current flowing through the circuit will be maximum when the impedance of the circuit is minimum. To do so, we vary the value of frequency \(\omega \) such that at a particular frequency \({\omega _0},\) we get, \({X_C} = {X_L},\) such that the impedance,

\(Z = \sqrt {{R^2} + {{\left( {{X_C}\,- {X_L}} \right)}^2}} = \sqrt {{R^2} + 0} = R\)

Thus, the current will be maximum, i.e., \(i = \frac{{{V_m}}}{R}\)

When the impedance of the series LCR circuit, \(Z = R,\) is equal to the resistance. Here this frequency \({\omega _0}\) is called the resonant frequency of the circuit.

For, \({X_C} = {X_L}\)

\(\frac{1}{{{\omega _0}C}} = {\omega _0}L\)

Or, \({\omega _0} = \frac{1}{{\sqrt {LC} }}\)

Thus, in a series LCR circuit, resonance occurs when the capacitive and inductive reactance are equal in magnitude but have a phase difference of \(180^\circ .\)

For series LCR circuit, the phase difference,

\(\emptyset = {\tan ^{ – 1}}\left( {\frac{{{X_C} – {X_L}}}{R}} \right)\)

For, \({X_C} = {X_L},\emptyset = 0,\) the circuit is in resonance.

\({X_C} > {X_L},\,\emptyset < 0,\) the circuit is predominately capacitive

\({X_C} < {X_L},\,\emptyset > 0,\) the circuit is predominately inductive

Summary

An LCR circuit is an electrical circuit consisting of an inductor (L), a capacitor (C), and a resistor (R). It is also known as a tuned or resonant circuit. The combination of these elements, when connected in series, is called a series LCR circuit.

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

The relation between the current and alternating voltage in a series LCR circuit: \({i_m} = \frac{{{V_m}}}{Z} = \frac{{{V_m}}}{{\sqrt {{R^2} + {{\left( {{X_C} – {X_L}} \right)}^2}} }}\;\)

The Phase relation between resistance, inductance, and capacitance in a series LCR circuit: \(\emptyset = {\rm{ta}}{{\rm{n}}^{ – 1}}\left( {\frac{{{X_C} – {X_L}}}{R}} \right)\)

In a series LCR circuit, resonance occurs when the capacitive and inductive reactance are equal in magnitude but have a phase difference of \(180^\circ ,\) and the frequency of the LCR circuit in this condition is called resonant frequency, \({\omega _0} = \frac{1}{{\sqrt {LC} }}.\)

FAQs on AC Voltage Applied to a Series LCR Circuit

Q.1. What is another name for the LCR circuit?
Ans: LCR circuit is also called a resonant circuit or tuned circuit.

Q.2. What is the resonance condition for the series LCR circuit?
Ans: At resonance, the capacitive and inductive reactance are equal and out by phase by \(180^\circ\).

Q.3. What is the impedance of the series LCR circuit?
Ans: The expression for impedance of series LCR circuit, \(Z = \sqrt {{R^2} + {{\left( {{X_C}\,- {X_L}} \right)}^2}} \).

Q.4. When is the applied voltage in phase with the instantaneous current?
Ans: For \(\theta = 0^\circ \): The applied voltage and instantaneous current are in phase.

Q.5. Write the expression for resonance frequency.
Ans: The resonance frequency, \({\omega _0} = \frac{1}{{\sqrt {LC} }}\).