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

    By Safeer PP
    31 May'22  7 min read
    NEET Previous Year Papers’ Analysis: LCR Circuits, Concepts And Solutions
    Synopsis

    Alternating current is an important chapter to be studied for the National Eligibility Cum Entrance Test (NEET). The chapter belongs to NCERT syllabus of Class 12 Physics. In the last six years’ paper of NEET there were eight questions from Alternating Current. Out of these eight questions, five were from the topic LCR circuits. In the JEE Main paper also the questions related to LCR circuits are frequently asked. Read on to understand the concepts covered under the topic LCR circuits and the NEET questions asked from RLC circuits for the last six years.

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

    Alternating current is an important chapter to be studied for the National Eligibility Cum Entrance Test (NEET). The chapter belongs to NCERT syllabus of Class 12 Physics. In the last six years’ paper of NEET there were eight questions from Alternating Current. Out of these eight questions, five were from the topic LCR circuits. In the JEE Main paper also the questions related to LCR circuits are frequently asked. Read on to understand the concepts covered under the topic LCR circuits and the NEET questions asked from RLC circuits for the last six years.

    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.

    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

    1653630750417

    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?

    1. L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m

    2. L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b

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

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

    Option c) is the right answer.

    Phasor Diagram Of Series RLC Circuit

    Phasor diagram for VL>VC

    1653630753171

    Phasor diagram for VC>VL

    1653630754041

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

    Resultant Voltage Of LCR Series Circuit

    1653630747928

    Question (JEE Main 2014)

    When the RMS voltages VL, VC and VR 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 VL: VC: VR = 1:2:3. If the RMS voltage of the AC source is 100 V, then VR (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

    Impedance Z Of LCR Series Circuit

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

    As XL=⍵L and XC=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.

    1653630748514

    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

    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

    1653630747576

    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 VS=Vmsinωt, the current I=Imsinωt.

    • For a purely inductive circuit, the instantaneous voltage leads the current by 900. That is if VS=Vmsinωt, I=Imsin(ωt-?/2)

    • For a purely capacitive circuit, the instantaneous voltage lags the current by 900. That is if VS=Vmsinωt, IS=Imsin(ωt+?/2)

    • For a series RLC circuit if source voltage VS=Vmsinωt, IS=Imsin(ωt-ɸ) and the value of ɸ may be positive or negative depending on the value of inductive and capacitive reactance. If XL>XC, the current lags behind the voltage. If XC>XL current leads to the voltage.

    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(XC=XL). The frequency at which XC=XL is known as resonant frequency ⍵0. At ⍵0, XC-XL=0, impedance becomes minimum and current I become maximum and Imax=Vmax/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 Vm=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=Vrms / Irms

    \\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 =Po/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

    1. A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistanceR= 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

    2. In a series LCR circuit, the inductive reactance (XL) is 10\Omegaand the capacitive reactance (XC) is 4\Omega. The resistance (R) in the circuit is 6\Omega. What is the power factor of the circuit?

    3. 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.

    • Quantity
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