Spread the love

Series LCR Circuit | A Series LCR Circuit Connected To An AC Source

Series LCR Circuit | A Series LCR Circuit Connected To An AC Source :- In the figure below, a pure resistor , an ideal capacitor $C$ , and a pure inductor $L$ are connected in series with an alternating source.

(i) Solution Of Series LCR Circuit Using Phasor Diagram

Let the applied alternating voltage be :

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ …..(1)

Since R, L, and C are connected in series, the magnitude and phase of the current flowing through all three components will be the same at any instant. Let the alternating current flowing through the circuit at any instant be :

$\displaystyle I={{I}_{0}}sin\text{ }\omega t$ …..(2)

According to the above figure, the potential difference across the resistor (R) will be in phase with the alternating current, therefore,

$\displaystyle {{V}_{R}}=IR=\left( {{I}_{0}}sin\text{ }\omega t \right)R=\left( {{I}_{0}}R \right)sin\text{ }\omega t={{V}_{0R}}sin\text{ }\omega t$ …..(3)

The potential difference across the inductor (L) will lead the current by 90° (π/2 radians) :

$\displaystyle {{V}_{L}}={{V}_{0L}}sin\left( \omega t+\frac{\pi }{2} \right)=\left( {{I}_{0}}{{X}_{L}} \right)sin\left( \omega t+\frac{\pi }{2} \right)$ …..(4)

And the potential difference across the capacitor (C) will lag the current by 90° (π/2 radians) :

$\displaystyle {{V}_{C}}={{V}_{0C}}sin\left( \omega t-\frac{\pi }{2} \right)=\left( {{I}_{0}}{{X}_{C}} \right)sin\left( \omega t-\frac{\pi }{2} \right)$ …..(5)

Impedance of Series LCR Circuit and the Peak Value of Alternating Current

In the figure below, the various voltages are represented by vectors :

Here, the electric current is represented along the X-axis, and the vectors $\displaystyle {{\overrightarrow{V}}_{OR}}$ , $\displaystyle {{\overrightarrow{V}}_{OL}}$ and $\displaystyle {{\overrightarrow{V}}_{OC}}$ are represented by $\displaystyle \overrightarrow{OA}$ , $\displaystyle \overrightarrow{OB}$ and $\displaystyle \overrightarrow{OC}$ , respectively.

Since the potential differences across L and C, denoted by $\displaystyle {{\overrightarrow{V}}_{OL}}$ and $\displaystyle {{\overrightarrow{V}}_{OC}}$ , have a phase difference of 180°, their resultant $\displaystyle \left( {{\overrightarrow{V}}_{OL}}-{{\overrightarrow{V}}_{OC}} \right)$ is represented along $\displaystyle \overrightarrow{OD}$ . (It is assumed that $\displaystyle \left( {{\overrightarrow{V}}_{OL}}>{{\overrightarrow{V}}_{OC}} \right)$ )

Now in ΔOAP,

$\displaystyle \overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}$

$\displaystyle \Rightarrow \overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{OD}$

$\displaystyle \Rightarrow {{\overrightarrow{E}}_{0}}={{\overrightarrow{V}}_{OR}}+\left( {{\overrightarrow{V}}_{OL}}-{{\overrightarrow{V}}_{OC}} \right)$ …..(6)

By Pythagoras theorem in ΔOAP,

$\displaystyle {{\left( OP \right)}^{2}}={{\left( OA \right)}^{2}}+{{\left( AP \right)}^{2}}$

$\displaystyle \Rightarrow {{\left( OP \right)}^{2}}={{\left( OA \right)}^{2}}+{{\left( OD \right)}^{2}}$ …..(7)

Using equations (6) and (7),

$\displaystyle {{\left( {{E}_{0}} \right)}^{2}}={{\left( {{V}_{0R}} \right)}^{2}}+{{\left( {{V}_{0L}}-{{V}_{0C}} \right)}^{2}}$

$\displaystyle \Rightarrow {{\left( {{E}_{0}} \right)}^{2}}={{\left( {{I}_{0}}R \right)}^{2}}+{{\left( {{I}_{0}}{{X}_{L}}-{{I}_{0}}{{X}_{C}} \right)}^{2}}$

$\displaystyle \Rightarrow {{\left( {{E}_{0}} \right)}^{2}}=I_{0}^{2}\left[ {{\left( R \right)}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}} \right]$

$\displaystyle {{E}_{0}}={{I}_{0}}\sqrt{\left[ {{\left( R \right)}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}} \right]}$ …..(8)

$\displaystyle {{I}_{0}}=\frac{{{E}_{0}}}{\sqrt{\left[ {{\left( R \right)}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}} \right]}}$ …..(9)

In equation (9), the quantity $\displaystyle \sqrt{\left[ {{\left( R \right)}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}} \right]}$ represents the obstruction produced in the path of the electric current due to the series LCR circuit. This quantity is called the impedance of the series LCR circuit. Impedance is denoted by , where …

$\displaystyle Z=\frac{{{E}_{0}}}{{{I}_{0}}}=\sqrt{\left[ {{\left( R \right)}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}} \right]}$ …..(10)

The S.I. unit of impedance is ohm (Ω).

The phase difference between the resultant electromotive force (E₀) and the current (I₀)

In ΔOAP,

$\displaystyle tan\phi =\frac{AP}{OA}=\frac{{{V}_{0L}}-{{V}_{0C}}}{{{V}_{0R}}}=\frac{{{I}_{0}}{{X}_{L}}-{{I}_{0}}{{X}_{C}}}{{{I}_{0}}R}=\frac{{{X}_{L}}-{{X}_{C}}}{R}$

$\displaystyle \therefore \phi =ta{{n}^{-1}}\left[ \frac{{{X}_{L}}-{{X}_{C}}}{R} \right]$ …..(11)

Thus, in a series LCR circuit, the resultant voltage can be represented as follows :

$\displaystyle E={{E}_{0}}sin\left( \omega t+\phi \right)$ …..(12)

Now, three situations arise here :

(a) When $\displaystyle {{X}_{L}}={{X}_{C}}$

$\displaystyle \therefore tan\phi =0$

$\displaystyle \Rightarrow \phi =0$

In this condition, the resultant reactance (X) of the circuit becomes zero, and the series LCR circuit behaves like a purely resistive circuit. In this situation, the resultant voltage (E₀) and current (I₀) are in the same phase, and this condition is also known as the condition of resonance.

(b) When $\displaystyle {{X}_{L}}>{{X}_{C}}$

$\displaystyle \therefore tan\phi =+ve$

$\displaystyle \Rightarrow \phi =+ve$

In this situation, the resultant reactance (X) of the circuit will be inductive reactance, and the series LCR circuit will behave like an inductance-dominated circuit. In this condition, the resultant voltage (E₀) leads the current (I₀) by a phase angle Φ, where 0°< Φ < 90°.

$\displaystyle \therefore tan\phi =-ve$

$\displaystyle \Rightarrow \phi =-ve$

In this situation, the resultant reactance (X) of the circuit will be capacitive reactance, and the series LCR circuit will behave like a capacitance-dominated circuit. In this condition, the resultant voltage (E₀) lags behind the current (I₀) by a phase angle Φ, where 0°> Φ > – 90°.

In a series LCR circuit, only the resistance is a component whose value does not depend on frequency, while all other components ( X_L, X_C, X, Z ) have values that depend on frequency, as shown in the figure below :

(ii) Analytical Solution of Series LCR Circuit

Suppose a pure resistance R, an ideal capacitor C, and a pure inductor L are connected in series with an alternating voltage source.

Let the applied alternating voltage be :

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ …..(13)

At any time , suppose…

charge on the plates of the capacitor = q

the value of the current in the circuit = I

$\displaystyle \frac{dI}{dt}$ = rate of change of current in the circuit

potential difference across the capacitor = $\displaystyle \frac{q}{C}$

potential difference across the inductor = $\displaystyle L\frac{dI}{dt}$

potential difference across the resistance = IR

Hence, from Kirchhoff’s law,

$\displaystyle L\frac{dI}{dt}+RI+\frac{q}{C}={{E}_{0}}sin\omega t$ …..(14)

Because the applied voltage is alternating, the current flowing in the circuit will also be alternating. The frequency of this current will be the same as that of the applied alternating voltage, but its amplitude and phase will be different. Let the current lag behind the voltage by a phase angle Φ, and let its amplitude be I₀. Therefore, the current in the circuit can be written as :

$\displaystyle I={{I}_{0}}sin\left( \omega t-\phi \right)$ …..(15)

then

$\displaystyle \frac{dI}{dt}=\omega {{I}_{0}}cos\left( \omega t-\phi \right)$

and

$\displaystyle q=\int{Idt=}\int{\left[ {{I}_{0}}sin\left( \omega t-\phi \right) \right]dt=}\frac{-{{I}_{0}}}{\omega }cos\left( \omega t-\phi \right)$

Putting the values of q, I And dI/dt in equation (14),

$\displaystyle L\left[ \omega {{I}_{0}}cos\left( \omega t-\phi \right) \right]+R\left[ {{I}_{0}}sin\left( \omega t-\phi \right) \right]+\frac{\left[ \frac{-{{I}_{0}}}{\omega }cos\left( \omega t-\phi \right) \right]}{C}={{E}_{0}}sin\omega t$

$\displaystyle \Rightarrow \omega L{{I}_{0}}\left[ cos\left( \omega t-\phi \right) \right]+R{{I}_{0}}\left[ sin\left( \omega t-\phi \right) \right]-\frac{{{I}_{0}}}{\omega C}\left[ cos\left( \omega t-\phi \right) \right]={{E}_{0}}sin\omega t$

$\displaystyle \Rightarrow \omega L{{I}_{0}}\left[ cos\omega tcos\phi +sin\omega tsin\phi \right]+R{{I}_{0}}\left[ sin\omega tcos\phi -cos\omega tsin\phi \right]-\frac{{{I}_{0}}}{\omega C}\left[ cos\omega tcos\phi +sin\omega tsin\phi \right]={{E}_{0}}sin\omega t$

Comparing the coefficients of cos ωt on both sides of the above equation, we get

$\displaystyle \omega L{{I}_{0}}cos\phi -R{{I}_{0}}sin\phi -\frac{{{I}_{0}}}{\omega C}cos\phi =0$

$\displaystyle \Rightarrow R{{I}_{0}}sin\phi =\left( \omega L-\frac{1}{\omega C} \right){{I}_{0}}cos\phi$

$\displaystyle \Rightarrow tan\phi =\frac{\left( \omega L-\frac{1}{\omega C} \right)}{R}$

$\displaystyle \Rightarrow \phi =ta{{n}^{-1}}\left[ \frac{\left( \omega L-\frac{1}{\omega C} \right)}{R} \right]$ …..(16)

Similarly, comparing the coefficients of sin ωt,

$\displaystyle \omega L{{I}_{0}}sin\phi +R{{I}_{0}}cos\phi -\frac{{{I}_{0}}}{\omega C}sin\phi ={{E}_{0}}$

$\displaystyle \Rightarrow {{I}_{0}}=\frac{{{E}_{0}}}{Rcos\phi +\left( \omega L-\frac{1}{\omega C} \right)sin\phi }$

$\displaystyle \Rightarrow {{I}_{0}}=\frac{{{E}_{0}}}{R\left[ cos\phi +\frac{\left( \omega L-\frac{1}{\omega C} \right)}{R}sin\phi \right]}$

$\displaystyle \Rightarrow {{I}_{0}}=\frac{{{E}_{0}}}{R\left[ cos\phi +tan\phi sin\phi \right]}$

$\displaystyle \Rightarrow {{I}_{0}}=\frac{{{E}_{0}}cos\phi }{R}$

From equation (16),

$\displaystyle cos\phi =\frac{R}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}$

Therefore

$\displaystyle {{I}_{0}}=\frac{{{E}_{0}}}{R}\times \frac{R}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}$

$\displaystyle \therefore {{I}_{0}}=\frac{{{E}_{0}}}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}$ …..(17)

Therefore, the value of current in the circuit at any instant,

$\displaystyle I={{I}_{0}}sin\left( \omega t-\phi \right)=\frac{{{E}_{0}}}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}sin\left( \omega t-ta{{n}^{-1}}\left[ \frac{\left( \omega L-\frac{1}{\omega C} \right)}{R} \right] \right)$

Here, the quantity $\displaystyle \sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}$ is called the impedance (Z) of a series LCR circuit.

Note :-

The phase difference between inductive reactance and capacitive reactance is π radians.
An inductor is called a low-pass filter because it allows low-frequency signals to pass.
A capacitor is called a high-pass filter because it allows high-frequency signals to pass.

Example 1.

(NCERT Example 7.6)

A resistor of 200 Ω and a capacitor of 15 μF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.

Solution :

(a) Impedance of the circuit

$\displaystyle Z=\sqrt{{{R}^{2}}+X_{C}^{2}}=\sqrt{{{R}^{2}}+{{\left( \frac{1}{2\pi \nu C} \right)}^{2}}}$

$\displaystyle Z=\sqrt{{{\left( 200 \right)}^{2}}+{{\left( \frac{1}{2\pi \times 50\times 15\times {{10}^{-6}}} \right)}^{2}}}$

$\displaystyle Z=\sqrt{{{\left( 200 \right)}^{2}}+{{\left( 212.3 \right)}^{2}}}$

$\displaystyle Z=291.5\Omega$

Hence, the current flowing in the circuit,

$\displaystyle I=\frac{V}{Z}=\frac{220}{291.5}=0.755Amp.$

(b) Since the same current flows in a series circuit, therefore…

V_R = IR = 0.755 × 200 = 151 Volts

V_C = IX_C = 0.755 × 212.3 = 160.3 Volts

The algebraic sum of the voltages = V_R + V_C = 311.06 volts, which is greater than the source voltage. As we know, there is a phase difference of 90º between V_Rand V_C, so they cannot be added like ordinary numbers. Taking the phase difference into account, their resultant must be determined as vectors. Therefore,

$\displaystyle {{V}_{R+C}}=\sqrt{V_{R}^{2}+V_{C}^{2}}=\sqrt{{{\left( 151 \right)}^{2}}+{{\left( 160.06 \right)}^{2}}}$

$\displaystyle {{V}_{R+C}}=\sqrt{22801+26519.2}=\sqrt{48420.2}$

⇒ V_R+C = 220 Volts = Source voltage

Example 2.

A 50W, 100 V lamp is to be connected to a 200V, 50Hz AC supply. What value of capacitor must be connected in series with the lamp ?

Solution :

Here, the maximum potential difference across the lamp can be 100 V, so to connect it to a 200 V source, another component must be added in series with it. According to the question, this component will be a capacitor.

Lamp resistance,

$\displaystyle R=\frac{V_{s}^{2}}{P}=\frac{{{(100)}^{2}}}{50}=\text{200}\Omega$

Therefore, the maximum safe electric current that can flow through the lamp,

$\displaystyle I=\frac{V}{R}=\frac{100}{200}=\frac{1}{2}A$

When a capacitor is placed in series with the lamp and connected to a 200 V alternating current source, then

V = IZ

$\displaystyle \Rightarrow Z=\frac{V}{I}=\frac{200}{{}^{1}\!\!\diagup\!\!{}_{2}\;}=400\Omega$

In C-R circuit,

$\displaystyle Z=\sqrt{{{R}^{2}}+\frac{1}{{{(\omega C)}^{2}}}}$

$\displaystyle \Rightarrow {{R}^{2}}+\frac{1}{{{(\omega C)}^{2}}}=16\times {{10}^{4}}$

$\displaystyle \Rightarrow \frac{1}{{{(\omega C)}^{2}}}=16\times {{10}^{4}}-{{(200)}^{2}}=12\times {{10}^{4}}$

$\displaystyle \Rightarrow \frac{1}{\omega C}=\sqrt{12}\times {{10}^{2}}$

$\displaystyle \Rightarrow C=\frac{1}{100\pi \times \sqrt{12}\times {{10}^{2}}}F$

$\displaystyle \therefore C=\frac{100}{\pi \sqrt{12}}\mu F=9.2\mu F$

Example 3.

In the circuit shown in the figure, $\displaystyle V=50\sqrt{2}sin\text{ }100\pi t$ volts and the ammeter reading is 2 A. Determine the value of .

Solution :

Ammeter measures the rms value of current, so

$\displaystyle {{I}_{rms}}=2Amp.$

$\displaystyle {{V}_{0}}=50\sqrt{2}\text{ }Volt$

$\displaystyle \Rightarrow {{V}_{rms}}=\frac{{{V}_{0}}}{\sqrt{2}}=50\text{ }Volt$

Now

$\displaystyle {{V}_{rms}}={{I}_{rms}}{{X}_{L}}$

$\displaystyle \Rightarrow {{X}_{L}}=\frac{{{V}_{rms}}}{{{I}_{rms}}}=\frac{50}{2}=25\Omega$

$\displaystyle \therefore L=\frac{{{X}_{L}}}{\omega }=\frac{25}{100\pi }=\frac{1}{4\pi }henry$

Example 4.

A 100 Ω resistor, an inductor of henry, and a capacitor of μF are connected in series to an alternating source of 200 V and 50 Hz. Find :-

(i) Total reactance (ii) Total impedance (iii) The value of the current in the circuit (iv) Phase difference between voltage and current (v) Susceptance

Solution :

Given :-

$\displaystyle \begin{array}{l}{{E}_{rms}}=200V\\\nu =50Hz\\R=100\Omega \\L=\frac{2}{\pi }H\\C=\frac{100}{\pi }\mu F=\frac{{{10}^{-4}}}{\pi }F\end{array}$