AC Voltage Applied to a Series LCR Circuit: Equation and Derivation

Education Journalist | Study Abroad Lead

LCR circuit, which is also known as tuned or resonant circuit, refers to an electrical circuit that includes an inductor L, a capacitor C and a resistor R which are connected in series so that the same amount of current flows in the circuit. Here, an alternating current or AC generator or dynamo can be used as AC voltage source.

Table of Content

AC voltage applied to a series of LCR Circuit
What is a series of LCR Circuits?
What is an Alternating Current?
How does a series of LCR Circuits work?
Previous Year Questions
Sample Questions

Keyterms: LCR Circuit, AC Generator, Alternating Current, Inductor, Electrical Circuit, Resistor, Current, Circuit, AC Voltage, Electrical Current

AC voltage applied to a series of LCR Circuit

[Click Here for Sample Questions]

It is common knowledge that an electrical current is required to make most of our daily appliances work. But are we ever curious to know how electrical current makes our appliances work? The most common form of electrical circuit found in most of the devices is the LCR Circuit and in this section, we will try to understand how a LCR Circuit works when AC Voltage (Alternating Current) is applied. But let's first understand how a series of LCR Circuits works.

AC circuit in an LCR series

AC circuit in an LCR series

Also Read:

Related Articles
RMS value of Ac	Average power consumption (Watts)	Power factor of AC Circuit
Impedence of the circuit	Calculating the value of Current in the circuit	Calculating the power dissipated

What is a series of LCR Circuits?

[Click Here for Previous Year Questions]

LCR Circuit is formed when a battery or any source with constant voltage is connected across a resistor as it then develops current in it. This current has a unique direction as it flows from the battery’s negative terminal to the positive terminal. The magnitude of the flowing current remains constant as well.

What is an Alternating Current?

[Click Here for Sample Questions]

Alternating Current can be defined as the periodic or alternate change in the direction of the flow of current through a resistor. AC Generators or an AC Dynamo are good examples of an AC Voltage source.

How does a series of LCR Circuits work?

[Click Here for Previous Year Questions]

We will understand the working of a series of LCR circuit with the help of the diagram given below:

Here, R= Resistor

L= Inductor

C= Capacitor

V= Voltage

The resistor, capacitor and the inductor are connected in a series to an AC voltage V, and the voltage is given by:

V=v_msinωt

In the above equation, v_m is the amplitude of the voltage and ω is the frequency of the voltage.

If we apply charge q on the capacitor and current i at time t, then by Kirchhoff’s loop rule, we get:

Ldi/dt+iR+qC=v

Here, R is the resistance of the resistor, q is the charge of the capacitor, i is the current in the circuit and C is the capacitance of the capacitor. To determine the instantaneous current in the circuit, we can follow the analytical solution given below:

For i,

i=dq/dt

We can also write this as,

di/dt=d²q/dt²

So, if we write the equation above in terms of charge ‘q’ running through the circuit, we get,

Ld²q/dt²+Rdq/dt+qC=v_msinωt

In order to solve the equation above, we can assume the solution given by:

q=q_msin(ωt+θ)

We can also write this equation as,

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

and as,

\(\frac{d^2q}{dt^2} = q_m \omega^2 cos (\omega t + \theta)\)

If we substitute these above values in the question of voltage, then we get,

q_mω [ R cos(ωt + θ) + (X_c – X_L) sin (ωt + θ)] = v_m sin ωt

In the above equation we have substituted the value of XC with 1/ωC and XL by ωL.

From what we have studied before, we know that,

\(z = \sqrt{R^2 = (X_c^1 - X_L^1)^2}\)

Therefore, if we substitute this above value in the equation above, we will get,

\(\frac{qm \omega}{Z} [ \frac{R}{Z} cos { (\omega t + \theta)} + (\frac{X_c - X_L}{Z})sin(\omega t + \theta)] = v_m sin \omega t\)

Substituting further,

\(\frac {R}{Z} = cos \phi, \frac{x_c - x_L}{z} = sin \phi,\)

we get,

\(\phi = tan ^{- 1} \frac{X_c - X_L}{R}\)

q_mωZ [cos(ωt + θ – Φ] = v_m sin ωt

If we compare two sides of the equation, then we can write it as,

q_mωZ = v_m = i_mZ

and further,

θ – Φ = – \(\frac{\pi}{2}\)

θ = – \(\frac{\pi}{2}\) + Φ

Thus, for the current in the circuit, we can write the equation as,

i = \(\frac{dq}{dt}\) = q_m ω cos (ωt + θ) = i_m cos (ωt + θ) = i_m sin (ωt + θ)

Also Read:

Related Articles
Time Constant Formula	Reactance and Impedance	Phasor Representation AC
Alternating Current Important Questions	Alternating Current Ncert Solutions	RMS value alternating current
Ohm’s Law MCQs	Electric Current Important Questions	Wheatstone Bridge MCQs
Ohm’s Law Important Questions	Meter Bridge MCQs	Wheatstone Bridge Important Questions
Potentiometer MCQs	Meter Bridge Important Questions	Electric Current MCQs

Previous Year Questions

Sample Questions

Ques. In a series LCR circuit, the voltage across an inductor, a capacitor and a resistor are 30 V, 30 V and 60 V respectively. Find out the phase difference between the applied voltage and the current in the circuit. (1 mark)

Ans. The phase difference between the applied voltage and the current in the series LCR circuit is,

LCR Circuit

Ques. The figure given below shows the series LCR circuit connected to a variable frequency 200 V source with L = 50 mH, C = 80 μF and R = 40 ?.
What is the source frequency which derives the current in resonance?
The quality factor Q of the circuit. (All India 2014C)

Ans. As given, L = 50mH = 50 x 10^-3 H

C = 80 μF = 80 x 10-6 F

R = 400 Π

V = 200 V

(i) In the LCR, the resonant angular frequency is,

LCR Circuit

Therefore, actual or source frequency is,

LCR Circuit

(ii) The quality factor

Quality Factor

Ques: In a series LCR circuit, derive the condition under which (i) the impedance of the circuit is minimum and (ii) wattless current flows in the circuit. (Foreign 2014)

Ans.

The impedance of a circuit LCR is,

LCR Circuit

When ωL = 1/ωC, Z will be minimum which means, when circuit is under resonance.

Hence, Z will be minimum and equal to R.

Average power dissipated through the series LCR circuit is,

LCR Circuit

Here, E_^v is the rms value of alternating voltage,

L_v is the rms value of alternating current,

Φ is the phase difference between current and voltage

For the wattless current, the power dissipated through the circuit must be zero.

LCR Circuit

Therefore, the condition for the wattless current is that the phase difference between the current and the circuit is purely capacitive and inductive.

Ques: Calculate the quality factor of a series LCR circuit having L = 2.0 H, C = 2μF and R = 10 ?. Mention the significance of quality factors in LCR circuit. (Foreign 2012)

Ans. Given,

L = 2.0 H

C = 2μF = 2 x 10^-6 μF

R = 10 Ω

Now, Q factor,

LCR Circuit

The quality factor can also be defined as,

Q = 2 Πf x energy stored/ power loss.

Ques. Find out the value of the phase difference between the current and the voltage in the series LCR circuit in the figure below. (All India 2015)

Ans. From the figure,

LCR Circuit

As tan Φ is negative X_L < X_C, the voltage lags behind the current.

Ques. (a) In a series LCR circuit which is connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot that shows its variation with frequency of the ac source.
(b) What is the phase difference between the voltages across the inductor and the capacitor at resonance in the LCR circuit?
(C) Explain why, When an inductor is connected to a 200 V de voltage, a current of 1A flows through it. Again, when the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Also calculate the self inductance of the inductor. (CBSE 2019)

Ans. In a series LCR circuit connected across an ac source with variable frequency.

Voltage of the ac source is V = V_m sin ?t

And, current is I = I_m sin ?t

The voltage drop across resistor R is,

V_R = I_m R

LCR Circuit

The voltage is in the same phase of the current. Now, the voltage drop across the inductor is,

V_L = I_m X_L

Here, X_L is the inductance, L X L = ω. Now the voltage leads the current by Π/2.

The voltage drop across capacitor is,
V_C = I_m X_C

Here, is the capacitive reactance, XC = 1/ωc

Again, the voltage leads the current by Π/2.

LCR Circuit

The resulting voltage will be the vector sum of all voltage which is represented by OF.

LCR Circuit

Thus,

As we know, V_m /I_m = z, therefore,

Which is the required expression for impedance.

Variation of impedance with frequency- at resources frequency, we have,

X_L = X_C = Z = R

LCR Circuit

(b) At resonance, the phase angle of circuit is zero,

LCR Circuit

Hence, impedance is,

Therefore, the voltage across the inductor and capacitor are the same at any instant and cancel out each other at resonance. Which is why the voltage drop across the LCR circuit is due to the voltage drop across the resistance R.

Hence, the phase difference between the voltages across the inductor and the capacitor at resonance is 180 degree.
(c) When an inductor is connected to 200 V dc voltage, it simply behaves as a resistor and there is no inductive reactance.

The resistance of the coil can be obtained by using Ohm’s law, V = IR.

Therefore, R = I/V

Given, 1 A current flows when the inductor is connected to 200 V dc.

Resistance

Hence,

When the same inductor is connected to 200 V, 50 Hz ac source, the inductive reactance gives rise to change in total impedance of the coil for which current changes.

Inductive reactance, X_L = W_L and frequency is 50 Hz

Therefore the angular frequency is,

The net impedance is,

By using Ohm’s law we get,

Therefore,

Substituting the values we get,

Impedance

(√12/Π) H is the self inductance of the inductor.

Ques. An ac source of voltage V = V₀ sin ωt is connected to a series combination of LCR. by using the phasor diagram, obtain the expression for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage and what is the circuit in this condition called? (Delhi 2016)

Ans.

LCR Circuit

From the above figure,

LCR Circuit

Here, the circuit in this condition is called a resonant circuit.

Ques: What is a resistor? (1 mark)

Ans: A resistor is a two-terminal device that is used to divide voltages or reduce current flow in a circuit. The S.I. unit used while measuring resistance is called ohms and is written as Ω.

Ques: What is an inductor? (1 mark)

Ans: An inductor is a passive instrument that is used to store energy in magnetic form when electricity is applied to the circuit. It is found in most of the electronic appliances and its S.I. unit is henry depicted by H.

Ques: What is a Capacitor? (1 mark)

Ans: It is a device that stores energy in the form of electrical energy. It has two terminals and is represented by the unit Farad. But Farad is a large unit so for solving textbook problems, smaller units of Farad called micro-farads (µF) or pico-farads pF is used.

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates

Check-Out:

PCMB Study Guides
NCERT Solutions for Class 12 Physics	Class 12 Physics Notes	Formulas in Physics
Topics for Comparison in Physics	Choice-based questions in physics	Topics in relation to physics
Physics Study Notes	Class 12 Physics Book PDF	Class 12 Physics Practicals
Important Derivations in Physics	SI units in Physics	Important Physics Constants and Units
Class 12 Physics Syllabus	Mendeleev's Periodic Table	NCERT Solutions for Class 12 Biology
NCERT Solutions for Class 12 English	NCERT Solutions for Class 11 Chemistry	Class 12 Chemistry Practicals
Class 12 Chemistry Notes	Important Chemical Reactions	Class 12 Maths Notes
Class 12 PCMB Notes	NCERT Class 12 Biology Book	NCERT Class 12 Maths Book
NCERT Class 12 Textbooks	NCERT Solutions for Class 11 Maths	NCERT Solutions for Class 11 Physics
NCERT Solutions for Class 12 Maths	NCERT Solutions for Class 11 Biology	NCERT Solutions for Class 11 English
NCERT Class 11 Chemistry Book	NCERT Class 11 Physics Book	NCERT Class 11 Biology Book
Class 11 Notes	Important Maths Formulas	Important Chemistry Formulas
Class 11 PCMB Syllabus	Chemistry Study Notes	Biology Study Notes
Periodic Table in Chemistry	Important Chemical Reactions	Comparison topics in Chemistry
Comparison Topics in Maths	Biology MCQs	Important Named Reactions
Maths MCQs	Comparison Topics in Biology	Chemistry MCQs

AC Voltage Applied to a Series LCR Circuit: Equation and Derivation

AC voltage applied to a series of LCR Circuit

What is a series of LCR Circuits?

What is an Alternating Current?

How does a series of LCR Circuits work?

Previous Year Questions

Sample Questions

CBSE CLASS XII Related Questions

1.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

2.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

4.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

Comments

SUBSCRIBE TO OUR NEWS LETTER

AC Voltage Applied to a Series LCR Circuit: Equation and Derivation

AC voltage applied to a series of LCR Circuit

What is a series of LCR Circuits?

What is an Alternating Current?

How does a series of LCR Circuits work?

Previous Year Questions

Sample Questions

CBSE CLASS XII Related Questions

1.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

2.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

4. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

2.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

4.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.