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.

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.

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.

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

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

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?

Option c) is the right answer.

Phasor diagram for V_{L}>V_{C}

Phasor diagram for V_{C}>V_{L}

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

**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:

From the given ratio

As X_{L}=⍵L and X_{C}=1/(⍵C), the impedance can be calculated as

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.

**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

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

**Instantaneous Power**

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

Using the trigonometric relation

**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

Where vi is the instantaneous power

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

Apparent power is measured in VA(Volt Ampere)

The power factor cosɸ can be calculated as-

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

Reactive power is measured in VAr (Volt Ampere reactive)

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

For a purely resistive circuit the instantaneous current is in phase with the instantaneous voltage. That is for V

_{S}=V_{m}sinωt, the current I=I_{m}sinωt.For a purely inductive circuit, the instantaneous voltage leads the current by 900. That is if V

_{S}=V_{m}sinωt, I=I_{m}sin(ωt-?/2)For a purely capacitive circuit, the instantaneous voltage lags the current by 900. That is if V

_{S}=V_{m}sinωt, I_{S}=I_{m}sin(ωt+?/2)For a series RLC circuit if source voltage V

_{S}=Vmsinωt, IS=I_{m}sin(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.

**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-

I_{0}/2=I_{0}sinωt

I_{0}/2I_{0}=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

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-

And the power factor at resonance=1

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

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

Q=⍵_{0}L/R. Also

Q=1/(⍵_{0}RC)

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=⍵_{0}L/R =⍵_{0}/2?⍵=⍵_{0}/(⍵_{2}-⍵_{1})

**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

**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.

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}

**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:

**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

So at 46 rad/s and 54 rad/s P=P_{0}/2

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

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance, an inductor of inductive reactance and an unknown capacitor. Find the value of capacitance to maximize the average power

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

An AC circuit has and L = 800 mH, connected in series. Find the quality factor of the circuit.

Both hard work and smart study are essential to get admission to prestigious engineering institutes like the IITs, NITs and IIITs. Qualifying for the JEE main is the first step toward this. Here in this article we are going to discuss brenoulli’s principle with help of previous years questions.

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