The National Eligibility Cum Entrance Test (NEET) Paper has questions from Physics, Chemistry and Biology. The weight of Physics is 25% of total marks. As NEET is highly competitive, each subject is important. Most of the NEET students find Physics difficult. But enough practice and clarity in concepts can make Physics easy for students. Each chapter of the Physics syllabus for NEET is important. Here is the analysis of the Class 12 chapter Alternating Current and explanations on concepts and questions of the topic, Alternating Current, the LCR circuits.

NEET Previous Year Papers’ Analysis: LCR Circuits, Concepts And Solutions

LCR Circuit(Image: Shutterstock)

In the last six years' paper of NEET there were 8 questions from AC circuits and out of these five questions were associated with the series LCR circuit. There were two questions from AC circuits in NEET 2021 in which one question was of LCR series circuit and another a combination of the concept of LCR circuit and series resonance. It shows that AC circuits and the concepts of LCR series circuits are important for NEET examination. JEE Main also has a number of questions from the LCR circuits. Let us have a look at the LCR series circuit.

LCR Series Circuit

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R, L and C are connected in series across an ac voltage source. The current may lead or lag the voltage depending on the values of resistance and inductive and capacitive reactance by some angle ɸ. The calculation of ɸ is discussed later.

Force Voltage Analogy

Applying Kirchoff's voltage Law to the given RLC series circuit we get

$\\V_s-V_R-V_L-V_C=0\\V_s=IR+L\frac{di}{dt}+\frac{1}{C}\int idt\\or\\V_s=L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{q}{C}------(1)$

The equation of damped or forced oscillator(Studied in the chapter, Oscillation of Class 11) is

$F=m\frac{d^2x}{dt^2}+B\frac{dx}{dt}+Kx------(2)$

Comparing (1) and (2) the inductance L is analogous to mass m, resistance R is analogous to frictional coefficient B and capacitance is analogous to the 1/K where K is the spring constant.

Force Voltage Analogy

Mechanical Quantity	Electrical Quantity
Force	Voltage
Velocity	Current
Displacement	Charge
frictional coefficient	Resistance
Mass	Inductance
Spring constant	1/Capacitance

Question (JEE Main 2020)

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be?

$L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m$
$L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b$
$L\leftrightarrow m,C\leftrightarrow \frac{1}{k},R\leftrightarrow b$
$L\leftrightarrow \frac{1}{b},C\leftrightarrow \frac{1}{m},R\leftrightarrow \frac{1}{k}$

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Option c) is the right answer.

Phasor Diagram Of Series RLC Circuit

Phasor diagram for V_L>V_C

1653630753171

Phasor diagram for V_C>V_L

1653630754041

ɸ is the phase difference(angle between the resultant voltage and the current)

Resultant Voltage Of LCR Series Circuit

Question (JEE Main 2014)

When the RMS voltages V_L, V_C and V_R are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio V_L: V_C: V_R = 1:2:3. If the RMS voltage of the AC source is 100 V, then V_R (in V) is close to

Solution:

$\\Given\\100=\sqrt{V_R^2+(V_L-V_C)^2}$

From the given ratio

$\\V_C=\frac{2}{3}V_R\ And \ V_L=\frac{1}{3}V_R$

$\\So,\\100=V_R\sqrt{1+\left ( \frac{1}{3}-\frac{2}{3} \right )^2}\\100=\frac{\sqrt{10}}{3}V_R\\V_R=30\sqrt{10}=94.86V$

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Impedance Z Of LCR Series Circuit

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

As X_L=⍵L and X_C=1/(⍵C), the impedance can be calculated as

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

Impedance Triangle

1653630749757

From the impedance triangle the angle phi can be calculated using trigonometric functions.

That is cosɸ=R/Z and cosɸ is known as the power factor.

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

$sin\phi=\frac{|X_L-X_C|}{Z}$

Question(JEE Main 2017)

A sinusoidal voltage of peak value 283 V and angular frequency (⍵) as 320 rad/s is applied to a series LCR circuit. Given that R=5 Ω, L=25 mH and C=1000 µF. The total impedance and phase difference between the voltage across the source and the current will be:

Solution:

Impedance

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

$X_{L} =\omega L = (2\pi f) L = 320\times25\times 10^{-3}\Omega=8\Omega$

$X_{C} = \frac{1}{\omega C} = \frac{1}{320\times10^{-3}} \Omega = \frac{1}{0.32}\Omega=3.125\Omega$

$Z=\sqrt{5^2+(8-3.125)^2}=6.98\Omega$

The phase difference between the voltage across the source and the current

$\phi=cos^{-1}\frac{R}{Z}=cos^{-1}\frac{5}{6.98}\approx 45^0$

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Power In An AC Circuit

Instantaneous Power

It is the product of instantaneous voltage and current. That is:

$\\p=V_msinwtI_msin(wt-\phi)\\$

Using the trigonometric relation

$sinasinb=\frac{1}{2}(cos(a-b)-cos(a+b))$

$p=\frac{1}{2}V_mI_m(cos\phi-cos(2wt-\phi))$

Average Power

Voltage and current are not constant in an AC circuit, they change with respect to time. So power also changes. So the average power over a period T is

$\\P_{avg}=\frac{1}{T}\int_{0}^{T}vidt$

Where vi is the instantaneous power

$\\P_{avg}=\frac{1}{T}\int_{0}^{T}vidt$

$\\P_{avg}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}V_mI_m[cos\phi-cos(2wt-\phi)]dt\\\int_{0}^{T}cos(2wt-\phi)dt=0\\So\ P_{avg}=\frac{V_mI_m}{2}cos\phi\\ P_{avg}=\frac{V_m}{\sqrt{2}}\frac{I_m}{\sqrt{2}}cos\phi=V_{rms}I_{rms}cos\phi$

Where cosɸ is known as the power factor.

Average power is measured in Watts. Average power is also known as active power or true power.

Also, average power can be written as

$\\P_{avg}=apparent \ power\times cos\phi$

Apparent power is measured in VA(Volt Ampere)

Power Factor

The power factor cosɸ can be calculated as-

$cos\phi=\frac{Average \ power}{Apparent\ power}$

Cosɸ can also be calculated from impedance triangle. Refer to the impedance triangle discussed above.

Reactive Power (Wattless Power)

$Reactive\ Power=V_{rms}I_{rms}sin\phi$

Reactive power is measured in VAr (Volt Ampere reactive)

Power Triangle

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From the power triangle ɸ can be computed using different trigonometric ratios.

cosɸ=Active Power/Apparent Power

sinɸ=Reactive Power/Apparent Power

tanɸ=Reactive Power/ Active Power

Some Points To Remember

For a purely resistive circuit the instantaneous current is in phase with the instantaneous voltage. That is for V_S=V_msinωt, the current I=I_msinωt.
For a purely inductive circuit, the instantaneous voltage leads the current by 900. That is if V_S=V_msinωt, I=I_msin(ωt-?/2)
For a purely capacitive circuit, the instantaneous voltage lags the current by 900. That is if V_S=V_msinωt, I_S=I_msin(ωt+?/2)
For a series RLC circuit if source voltage V_S=Vmsinωt, IS=I_msin(ωt-ɸ) and the value of ɸ may be positive or negative depending on the value of inductive and capacitive reactance. If X_L>X_C, the current lags behind the voltage. If X_C>X_L current leads to the voltage.

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Question: (JEE Main 2019)

An alternating voltage V(t)=220sin100?t volt is applied to a purely resistive load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is:

Solution-

I₀/2=I₀sinωt

I₀/2I₀=sinωt

½=sinωt

ωt=?/6

Given ω=100?

100?t=?/6, therefore t=1/600 or =1..67milli seconds. This is the time to reach half the peak value from zero.

Peak value occurs at ωt=?/2. So time taken to reach peak value from zero=1/200=5milli seconds.

The time taken from half of the peak to the peak value=5-1.67=3.3milli seconds

Series Resonance

Resonance is the characteristic of the RLC series circuit. Resonance occurs when the value of capacitive and inductive reactance are equal(X_C=X_L). The frequency at which X_C=X_L is known as resonant frequency ⍵₀. At ⍵₀, X_C-X_L=0, impedance becomes minimum and current I become maximum and I_max=V_max/R. And the value of resonant frequency is-

$w_0=\frac{1}{\sqrt{LC}}$

And the power factor at resonance=1

Frequency Response Or Resonant Curve

The curve represents the variation of current amplitude with respect to frequency.

1653630746608

The sharpness of resonance is measured by the quality factor Q.

Q=⍵₀L/R. Also

Q=1/(⍵₀RC)

As Q increases the sharpness also increases. From the frequency response, we can say that the Q factor of (i) is greater than the Q factor of (ii) as the sharpness of the first curve is more.

Suppose we choose a frequency other than ⍵0 such that the current magnitude is 1/√2 times the maximum value or power dissipated by the circuit is half of maximum value. There will be two such frequencies for which the current magnitude is 1/√2 times the maximum value, ⍵₁<⍵₀ and ⍵₂>⍵₀. The magnitude of ⍵₀-⍵₁=⍵₂-⍵₀ =?⍵. Here ⍵1 and ⍵2 are half power frequencies.

The difference ⍵₂-⍵₁ = 2?⍵ is known as the bandwidth and bandwidth 2?⍵= R/L. Therefore Q=⍵₀L/R =⍵₀/2?⍵=⍵₀/(⍵₂-⍵₁)

1653630752934

Past 6 Years NEET Questions On RLC Series Circuit

NEET 2016

An inductor 20 mH, a capacitor 50 ?F and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in the A.C circuit-

Solution

Power loss in the circuit is asked. We have to find the average power.

Given V_m=10, ⍵=340. To find the average power we need to find Im and cosɸ

$\\I_m=\frac{V_m}{Z}=\frac{10}{Z} \\cos\phi=\frac{R}{Z}\\P_{avg}=\frac{1}{2}V_mI_mcos\phi\\=V_m\times\frac{V_m}{2Z}\times\frac{R}{Z}=\frac{V_m^2R}{2Z^2}\\\\$

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

$=\frac{1}{2}\left [ \frac{100\times40}{1600+(340\times20\times10^{-3}-\frac{1}{340\times50\times10^{-6}})^2} \right ]\\=\frac{2000}{1600+(6.8-58.82)^2}\\=\frac{2000}{4306}=0.464W$

NEET 2018

An inductor 20 mH, a capacitor 100 ?F and a resistor 50 Ω are connected in series across a source of emf V = 10 sin 314 t. The power loss in the circuit is

Solution: 0.79W

This question is similar to the NEET 2016 question and the values only differ. Try to solve it yourself.

NEET 2021

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in the figure.

1653630751867

The potential difference across L, C and R is 40 V, 10V and 40V, respectively. The amplitude of the current flowing through the LCR series circuit is 10√2 A. The impedance of the circuit is-

Solution:

Z=V_rms / I_rms

$\\Z=\frac{\sqrt{V_R^2+(V_L-V_C)^2}}{I_{rms}}\\=\frac{\sqrt{40^2+30^2}}{10\sqrt{2}/\sqrt{2}}=5\Omega$

NEET 2020

A series of LCR circuits is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is ?/3 If instead C is removed from the circuit, the phase difference is again ?/3 between current and voltage. The power factor of the circuit is

Solution:

$\begin{aligned} &\text { For only } \mathrm{RC}:\phi_{\mathrm{RC}}=\frac{\pi}{3}\\ &\tan \left(\frac{\pi}{3}\right)=\frac{\mathrm{X}_{\mathrm{c}}}{\mathrm{R}} \Rightarrow \mathrm{X}_{\mathrm{c}}=\sqrt{3} \mathrm{R}\\ &\text { For only } \mathrm{RL}:\phi_{\mathrm{RL}}=\frac{\pi}{3}\\ &\tan \left(\frac{\pi}{3}\right)=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \Rightarrow \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}\\ &\text { For RLC:- }\\ &\tan (\phi)=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=\frac{\sqrt{3} \mathrm{R}-\sqrt{3} \mathrm{R}}{\mathrm{R}}=0 \Rightarrow \phi=0 \Rightarrow \cos \phi=1 \end{aligned}$

NEET 2021

A series LCR circuit containing 5.0 H inductor 80 capacitor ?F and 40Ω resistor is connected to a 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:

(1) 25 rad/s and 75 rad/s (2) 50 rad/s and 25 rad/s (3) 46 rad/s and 54 rad/s (4) 42 rad/s and 58 rad/s

Solution:

Power =P_o/2 for half power frequencies. 2?⍵ the Bandwidth of resonance curve= R/L

$\\w_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5\times80\times 10^{-6}}}=50rad/sec.\\2\Delta w=\frac{R}{L}\\\Delta w=40/10=4rad/sec$

$\\w_0-w_1=4\\w_1=50-4=46rad/s$

$\\w_2-w_0=4\\w_2=50+4=54rad/s$

So at 46 rad/s and 54 rad/s P=P₀/2

So option (3) 46 rad/s and 54 rad/s is right

Practice Question

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance $R= 3k\Omega$ , an inductor of inductive reactance $X_{L}= 250\pi\, \Omega$ and an unknown capacitor. Find the value of capacitance to maximize the average power
In a series LCR circuit, the inductive reactance (XL) is 10 $\Omega$ and the capacitive reactance (XC) is 4 $\Omega$ . The resistance (R) in the circuit is 6 $\Omega$ . What is the power factor of the circuit?
An AC circuit has $R=100 \Omega, C= 2\mu F$ and L = 800 mH, connected in series. Find the quality factor of the circuit.

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A Veterinarian is responsible for diagnosing, treating, and preventing diseases in animals. The individual performs surgeries, guides nutrition, and provides animal care. A Bachelor’s in Veterinary Science (B.Vsc.) is a mandatory degree. The profession brings together medical knowledge and a strong commitment to animal welfare.

5 Jobs Available

Speech Therapist

A Speech Therapist (Speech-Language Pathologist) diagnoses and treats speech, language, communication, and swallowing disorders across all ages. They work in hospitals, schools, clinics, and more. Becoming an SLP requires a master’s degree, clinical training, and certification. With rising demand, the career offers rewarding opportunities in therapy, education, and research.

4 Jobs Available

Gynaecologist

A gynaecologist is a medical specialist in women’s reproductive health, handling issues like menstruation, fertility, pregnancy, and childbirth. They perform exams, surgeries, and offer family planning services. To become one, students must complete MBBS and postgraduate training. Gynaecologists work in hospitals or clinics and are in high demand, with salaries growing significantly with experience.

4 Jobs Available

Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available

Oncologist

An oncologist is a medical doctor who diagnoses and treats cancer using chemotherapy, radiation, surgery, and other therapies. They work with a team to create treatment plans tailored to each patient. Specialisations include medical, surgical, radiation, pediatric, gynecologic, and hematologic oncology. Becoming an oncologist in India requires an MBBS and postgraduate studies in oncology.

3 Jobs Available

Biochemist

A biochemist studies the chemical processes within living organisms, combining chemistry and biology. They conduct experiments, analyse data, and develop products like drugs and vaccines. Biochemists work in labs, healthcare, research, and education. A degree in biochemistry or related fields is essential, with advanced roles often requiring higher degrees. They also ensure quality control and may teach or mentor others.

2 Jobs Available

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