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Introduction to Alternating Current


An alternating current (a.c.) is a current which continuously, changes in magnitude and periodically reverses in direction.`
i = I0 sin ωt = I0 sin (2π/T) t
Here I0 is the peak value of a.c.

Basic Parameters and Equations

(a) Current, I =I0 sin ωt
(b) Angular frequency, ω= 2πn   (n is the frequency of a.c.)
(c) I =I0 sin 2πnt

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

Mean Value of AC or DC Value of AC


Mean value and its calculation for half-cycle
Mean value of a.c. is that value of steady current which sends the same amount of charge, through a circuit, in same time as is done by a.c. in one half-cycle.
(Iav)half cycle = (2/π)I0
Thus, mean value of alternating current is 2/π times (0.637 times) its peak value.
(Vav)half cycle = (2/π) V0

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

Average value of AC over a complete cycle
Iav = 0
The average value of a.c. taken over the complete cycle of a.c.is zero.

Root Mean Square (RMS) value of AC and its calculation
Root mean square value of alternating current is defined as that value of steady current which produces same heating effect, in a resistance, in a certain time as is produced by the alternating current in same resistance in same time. The r.m.s value of a.c.is also called its virtual value.
Irms = I0/√2
Root mean square value of alternating current is I/√2 times (or 0.707 times) 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

(a) Inductive reactance
XL = ωL
Here, ω = 2πn, n being frequency of a.c.
L is the coefficient of self-inductance of coil.

(b) Capacitative reactance
Xc = 1/ωC
Here C is the capacity of the condenser

AC Circuits with Components

Capacitor in AC circuit

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

q = CV0sinωt
I = I0 sin(ωt +π/2)
V0 = I0/ωC
Xc = 1/ωC

Inductor in AC circuit

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

V= L(dI/dt) = LI0ω cosωt
I = (V0/ ωL) sinωt
Here, I0 = V0/ ωL
XL = ωL
And the maximum current, I0 = V0/XL

R-L circuit

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

I = ε/R [1-e-Rt/L]
V = ε e-Rt/L

Graph between I (amp) and t (sec):-

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

Graph between potential difference across inductor and time:-

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

L-C Circuit:-

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

f = 1/2π√LC
q = q0 sin (ωt+ϕ)
I = q0ωsin (ωt+ϕ)
ω = 1/√LC

The total energy of the system remains conserved,
½ CV2 + ½ Li2 = constant = ½ CV02 = ½ Li02

Series in C-R circuit:-

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

V = IZ
The modulus of impedance, |Z |= √R2+(1/ωC)2
The potential difference lags the current by an angle, ϕ = tan-1(1/ωCR)

Series in L-C-R Circuit:

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced

V = IZ
The modulus of impedance, |Z |= √[R2+(ωL-1/ωC)2]
The potential difference lags the current by an angle, ϕ = tan-1[ωL -1/ωC)/R]

Circuit elements with A.C

Revision Notes: Alternating Currents | Physics for JEE Main & Advanced


Resonance:-
(a) Resonance frequency:- fr = 1/2π√LC
(b) At resonance, XL = XC, ϕ = 0, Z = R(minimum), cosϕ = 1, sinϕ = 0 nad current is maximum (=E0/R)

Half power frequencies:-

(a) Lower,  f1 = fr – R/4πL    or      ω1 = ωr – R/2L
(b) Upper,  f2 = fr + R/4πL    or      ω2 = ωr + R/2L

Band width:- Δf = R/2πL   or   Δf = R/L

Quality Factor
(a) Q = ωr/Δω = ωrL/R
(b) As ω = 1/√LC, So Q ∝ √L, Q ∝1/R and Q ∝ 1/√C
(c) Q = 1/ωrCR
(d) Q = XL/R   or  Q = XC/R
(e) Q = fr/Δf   

At resonance, peak voltages are:-
(a) (VL)res = e0Q
(b) (VC)res = e0Q
(c) (VR)res = e0

Conductance, Susceptance, and Admittance


(a) Conductance, G = 1/R
(b) Susceptance, S = 1/X
(c) SL = 1/XL and SC = 1/XC = ωC
(d) Admittance, Y = 1/Z
(e) Impedance add in series while add in parallel

Power in AC Circuits

Circuit containing pure resistance:- Pav = (E0/√2)×(I0/√2) = Ev×Iv

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

Circuit containing impedance (a combination of R,L and C):-
Pav = (E0/√2)×(I0/√2) cosϕ = (Ev×Iv) cosϕ
Here cosϕ is the power factor.
(a) Circuit containing pure resistance, Pav = EvIv
(b) Circuit containing pure inductance, Pav = 0
(c) Circuit containing pure capacitance, Pav = 0
(d) Circuit containing resistance and inductance,
Z = √R2+(ωL)
cosϕ = R/Z = R/[√{R2+(ωL)2}]
(e) Circuit containing resistance and capacitance:-
Z = √R2+(1/ωC)2
cosϕ = R/Z = R/[√{R2+(1/ωC)2}]
(f) Power factor, cosϕ = Real power/Virtual power = Pav/ErmsIrms

Transformer


(a) Cp = Np (dϕ/dt) and es = Ns (dϕ/dt)
(b) ep/es = Np/Ns
(c) As, epIp = esIs, Thus, Is/Ip = ep/es = Np/Ns
(d) Step down:- e< ep, Ns< Np and Is> Ip
(e) Step up:- e>ep, Ns>Np and Is< Ip
(f) Efficiency, η = es Is/ ep Ip

AC Generator


e = e0 sin (2πft)
Here, e0 = NBAω

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