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 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 Current | Graph |
---|---|
Constant DC | |
Periodic DC | |
Variable DC | |
AC Analog | |
AC Digital |
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.
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.
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 = rms value/average value = (V0/√2)/ (2 V0/π) = π/2√2
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 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
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 potential difference across inductor and time
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 | Amplitude relation | Circuit quantity | Phase of V |
---|---|---|---|
Resistor | V0 = i0R | R | In phase with i |
Capacitor | V0 = i0XC | C | Lags i by 90° |
Inductor | V0 = i0XL | XL = wL | Leads i by 90° |
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
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)]
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 A | Column 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
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
*The article might have information for the previous academic years, please refer the official website of the exam.