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    JEE Main Study Notes for Alternating Current: Formulas, Circuit Diagrams and Study Tips

    Nikkil Visha Nikkil Visha
    Exams Prep Master

    Alternating Current is a very important and scoring topic for JEE Main candidates. This chapter is considered as one of the difficult chapters of physics as it involves various formulas, circuit diagrams, facts, etc. In order to score positive marks from this chapter, candidates need to put extra effort. Most of the time students get confused while solving the questions on circuits because students focus more on formulas and forget to go through the circuit diagrams. 

    • With good preparation, you can directly score 8 to 12 marks from this section. 

    • Remember to revise the topics like voltage applied to a resistor, AC voltage applied to an inductor, AC voltage applied to a capacitor, AC voltage applied to a series LCR circuit. Check JEE Main Physics Syllabus

    Students can check the notes of Alternating Current based on the latest syllabus below in this article so that you can prepare well for JEE Main.

    Must Read:


    Alternating Current

    Alternating Current changes its direction periodically whereas if the direction of current is constant then it is called direct current or D.C. 

    Let us consider a sinusoidal varying function 

    i = Im sin(ωt+Φ)

    Here, the maximum current i.e. the peak current is denoted by Ιm and ‘i’ is the instantaneous current. 

    The factor (ωt+Φ) is called phase and ω is termed as angular frequency. 

    In terms of frequency, ω = 2πf. Also frequency f = TimePeriod(T)1.

    Here I0 is the peak value of a.c.

    Current, I =I0 sin ωt

    Angular frequency, ω = 2πn (n is the frequency of a.c.)

    I =I0 sin 2πnt

    Type of CurrentGraph
    Constant DC
    Periodic DC
    Variable DC
    AC Analog
    AC Digital

    Mean value of A.C or D.C. value of A.C.

    Mean value of A.C or D.C. value of A.C.

    Mean value of a.c. is the value of steady current which sends the same amount of charge, through a circuit, in the same time as is done by a.c. in one half-cycle.

    Thus, mean value of alternating current is 2/π times i.e. 0.637 times of its peak value.


    Average value of A.C. over a complete cycle

    Average value of A.C. over a complete cycle

    It is the average of all the instantaneous values of alternating current and alternating voltages over 1 full cycle. The +ve half cycle is exactly equal to the – ve half cycle in symmetrical waves like voltage or current waveform.

    Iav = 0

    The average value of a.c. taken over the complete cycle of a.c.is zero.


    Root mean square value of A.C. or virtual value of A.C.

    The root mean square value of A.C. is the value of steady current that produces the same heating effect, in resistance, in a certain time as is produced by the alternating current in the same resistance at the same time. The r.m.s value of a.c.is also known as its virtual value.

    Irms = I0/√2

    Root mean square value of alternating current is I/√2 times or 0.707 times of the peak value of current.

    Similarly, Vrms= V0/√2

    Here V0 is the peak value of e.m.f.

    Form Factor

    Form Factor = rms value/average value = (V0/√2)/ (2 V0/π) = π/2√2


    Current Elements

    Here are various current elements are discussed below.

    Inductive reactance

    • XL = ωL

    • Here, ω = 2πn, n is the frequency of a.c.

    • L is the coefficient of self-inductance of the coil.

    Capacitative reactance

    • Xc = 1/ωC

    • Here C is the capacity of the condenser


    Capacitor & Inductor in A.C. Circuit

    Capacitor and Inductor in A.C. Circuit

    Capacitor in A.C Circuit

    • q = CV0 sinωt

    • I = I0 sin (ωt +π/2)

    • V0 = I0/ωC

    • Xc = 1/ωC

    Inductor in A.C. Circuit

    • VL = L(dI/dt) = LI0 ω cosωt

    • I = (V0/ ωL) sinωt

    • Here, I0 = V0/ωL

    • XL = ωL

    • And the maximum current, I0 = V0/XL


    R-L circuit & L-C Circuit

    R-L circuit and L-C Circuit

    R-L circuit

    • I = ε/R [1-e-Rt/L]

    • V = εe-Rt/L

    L-C Circuit

    • f = 1/2π√LC

    • q = q0 sin (ωt+ϕ)

    • I = q0ωsin (ωt+ϕ)

    • ω = 1/√LC

    • The total energy of the system remains conserved,

    • ½CV2+ ½Li2= constant = ½CV02= ½Li02


    Graph between I (amp) and t (sec)

    Graph between potential difference across inductor and time


    Series in C-R circuit and L-C-R Circuit

    Series in C-R & L-C-R Circuit

    Series in C-R circuit

    V = IZ

    The modulus of impedance, |Z|= √R2+(1/ωC)2

    The lag in the current due to potential difference can be represented by an angle,ϕ = tan-1(1/ωCR)

    Series in L-C-R Circuit

    V = IZ

    The modulus of impedance, 

    •  |Z|= √[R2+(ωL-1/ωC)2]

    •  ϕ = tan-1[ωL -1/ωC)/R]


    Circuit elements with A.C

    Circuit elementsAmplitude relationCircuit quantityPhase of V
    ResistorV0 = i0RRIn phase with i
    CapacitorV0 = i0XCCLags i by 90°
    InductorV0 = i0XLXL = wLLeads i by 90°

    Resonance, Bandwidth, & Quality Factor

    Resonance, Bandwidth, and Quality Factor

    Condition for resonance: The Amplitude of current in an RLC circuit is maximum for a given value of R and V if the impedance of the circuit is minimized. 

    Resonance occurs when XL = XC

    •  Resonance frequency:- fr= 1/2π√LC

    • At resonance, XL= XC

    • ϕ = 0, Z = R(minimum)

    • cosϕ = 1, sinϕ = 0 

    • current is maximum (=E0/R)

    Half power frequencies

    • Lower, f1 = fr – R/4πL or ω1 = ωr – R/2L

    • Upper, f2 = fr + R/4πL or ω2 = ωr + /2L

    Band width

    Δf = R/2πL or Δf = R/L

    Quality Factor

    • Q = ωr/Δω = ωrL/R

    • As ω = 1/√LC, So Q ∝ √L, Q ∝1/R and Q ∝ 1/√C

    • Q = 1/ωrCR

    • Q = XL/R or Q = XC/R

    • Q = fr/Δf

    At resonance, peak voltages are

    • (VL)res = e0Q

    • (VC)res = e0Q

    • (VR)res = e0

    Conductance, susceptance and admittance

    • Conductance, G = 1/R

    • Susceptance, S = 1/X

    • SL = 1/XL 

    • SC = 1/XC = ωC

    • Admittance, Y = 1/Z

    • Impedance add in series while add-in parallel


    Power in AC circuits

    A circuit containing pure resistance

     Iv and Ev are the virtual values of the current and e.m.f respectively.

    A circuit containing impedance (a combination of R,L, and C)

    Here cosϕ is the power factor.

    • Circuit containing pure resistance, Pav = EvIv

    • Circuit containing pure inductance, Pav = 0

    • Circuit containing pure capacitance, Pav = 0

    Circuit containing resistance and inductance,

    Z = √R2+(ωL)

    cosϕ = R/Z = R/[√{R2+(ωL)2}]

    Circuit containing resistance and capacitance

    Z = √R2+(1/ωC)2

    cosϕ = R/Z = R/[√{R2+(1/ωC)2}]

    Power factor

    cosϕ = Real power/Virtual power = Pav/ErmsIrms

    Transformer

    • Cp =Np(dϕ/dt) and es = Ns(dϕ/dt)

    • ep/es = Np/Ns

    • Efficiency, η = esIs/epIp

    AC Generator

    e = e0 sin (2πft)

    Here, e0 = NBAω

    Power Consumed

    Let the potential difference v=VA−VB=VMsinωt

    Let the electric current flowing through it be, i=Isin(ωt+Φ)

    Therefore, the instantaneous power consumed by the device is given by

    P=(Vmsinωt)[Imsin(ωt+Phi)]


    Previous Year Questions with Solutions

    Previous Year Questions with Solutions

    Linked comprehensive Question Type 1

    Statement: A circuit consists of a capacitor of Xc=30 ohm, a non-inductive resistor of 44 ohms and a coil of inductive resistance 90 ohms and resistance 36 ohms in series is connected to 200 V, and 60 Hz AC circuit

    Question. Find the Impedance (Z) of the circuit

    (a) 10 ohm

    (b) 100 ohm

    (c) 30 ohm

    (d) 40 ohm

    Solution: Given Xc=30 ohm, R1=44 ohm, XL=90 ohm, R2=36 ohm

    Impedance (Z) of the circuit is given by

    Z=√(R1+R2)2+(XL−XC)2

    Z=√(44+36)2+(90−30)2=100Ω

    Question: Find the current in the circuit?

    (a) 1 A

    (b) 3 A

    (c) 2 A

    (d) 5 A

    Solution: Now current in the circuit is given by I=VZ=200/100=2A

    Question: Find the Impedance of the coil

    (a) 97 ohm

    (b) 50 ohm

    (c) 90 ohm

    (d) None of these

    Solution: The impedance of the coil us given by

    Zc=√R22+X2L =97 ohm

    Question: Match the following

    Column AColumn B
    (A) VCoil(P) 88 V
    (B) VRes(Q) 194 V
    (C) VCap(R) 200 V
    (D) VCoil + VRes + VCap(S) 60 V
    -(T) No appropriate match

    Solution: Now VC=IXC=2*30=60 V, VR=IR=2*44=88V, VL=IZC=2*97=194 V

    VCoil + VRes + VCap= 60 +88+194=342 V

    So,

    A = Q

    B = P

    C = S

    D = T

    Linked comprehensive Question Type 2

    Statement: A circuit consists of a series combination of a 50mH inductor and a 20μF capacitor. The circuit is connected to an Alternating Current supply of 220V and 50 Hz .The circuit has the value of R=0

    Question: Which of the following is correct for the circuit?

    (a) I0=2.17 A ,Irms= 1.53 A

    (b) I0=5.1 A ,Irms= 3.6 A

    (c) V0=311 V ,Vrms=200 V

    (d) None of these

    Solution: The following quantities are given in the question

    L= 50mH=50X10-3 H

    C = 20μF=20X10-6F

    Vrms = 220V

    F = 50hz 

    or ω = 2πf = 100π rad/sec

    Now

    V0=√Vrms=311 V

    The peak current is given by for LC circuit

    I0=V0Z

    Where

    Z=ωL−1ωC

    Substituting the values from above

    I0 = -2.17 A

    So magnitude is

    I0 = 2.17 A

    Irms = I0√2= 1.53 A


    Tips and Tricks to Solve Questions

    Tips and Tricks to Solve Questions on Alternating Current

    • If the circuit diagram is given without any values then identify the variables from the question and mention all the values in that diagram so that you can easily find the solution. 

    • If the circuit diagram is not given then draw one if necessary. 

    • Impedance will be calculated in Ohms and current will be calculated in Amperes. 

    • Knowledge of the right units or dimensions is very important before staring the question in JEE Main 2020 examination. 

    • The phase of resistance is R is 0.

    • The phase of capacitive reactance is -90 degree

    • The phase of inductive reactance is 90 degree

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