AC1

aAlternating CurrentThe current or emf whose magnitude changes with time and direction reverses periodically is known as alternating current or alternating emf.A sinusoidal current and emf can be expressed as:a

I	=I0sin𝜔t…(i)
E	=E0sin𝜔t…(ii)

where I0 and E0 are their maximum value or peak value or amplitude of current and e.m.f respectively. Their angular frequency is called driving frequency and is given by𝜔=2𝜋

T=2𝜋fWhere T is time period and f is frequency of alternating current or e.m.f.Advantages1. Generation, transmission, and distribution of A.C. are more economical than that of D.C.2. AC is easily convertible into D.C.3. AC can be better controlled without any loss of electric power by using a choke coil.4. The alternating high voltage can be stepped down or stepped up easily by using a transformer.5. AC. can reach distant places without much loss of electric power by using a transformer.6. AC. machines are easier to use.Disadvantages1. C. cannot be used in the electrolysis process such as electroplating, electrotyping, and electro refining etc. where only D.C. is used.2. C. can be more dangerous than D.C. in terms of its attractive nature and also because its maximum value is 2 times its effective value.3. C. in a wire is not uniformly distributed through its cross-section. The a.c. density is much greater near the surface of the wire than that inside the wire. The a.c. density is much greater near the surface of the wire than that inside the wire. The concentration of a.c. near the surface of the wire is called the skin effect.4. Markings on the scales of A.C. meters are not equidistant for small measurement.

Peak Value RMS value of AC current and VoltageThe RMS value of AC current is equal to that amount of DC current which produces the same heating effect flowing through the same resistance for the same time.It is also known as virtual value of ac or Effective value of ac.Let us consider the alternating current is represented by,I=I0sin𝜔tIf this alternating current flows through a resistance R for small time dt, then small amount of heat produced which is given bya

dH	=I2Rdt=(I20sin2𝜔t)Rdt
or, dH	=I2oRsin2𝜔Tdt...........(i)

To obtain total amount of heat produced in one cycle, we have to integrate equation (i) from t=0 to t=T.a

H	=I2oRT∫0sin2𝜔tdt
	=I2oRT∫0(1-cos2𝜔t) 2dt
	=I20R 2T∫0(1-cos2𝜔t)dt
	=I20R 2[T∫0dt-T∫0cos2𝜔tdt]T
	=I20R 2[[t]T0-[sin2𝜔t 2𝜔]T0]
	=I20R 2[(T-0)-1 2𝜔[sin22𝜋 Tt]T0]
	=I20R 2[T-1 2𝜔(sin22𝜋 TT-sin0)]
or, H	=I20RT 2…(ii)

If Irms be the r.m.s. value or virtual value of a.c., then the amount of heat produced in the same resistance R in same time T is writtenH=I2rmsRT…(iii)From equation (ii) and (iii), we getI2rmsRT=I20RT

2∴Irms=I0

2=0.707I0Hence the r.m.s. value or virtual value or effective value of a.c. in 0.707 times the peak value of a.c. i.e 70.7% of the peak value of a.c.The virtual value of alternating current can be proved asEv=E0

2=0.707E0 Mean and average value of ACAlternating current is positive during the first half cycle and negative during another half cycle, so the mean average value of a.c. over one cycle is zero. We can find the mean or average value of a.c. over any half cycle.Mean or average value of alternating current is that value of steady current which sends the same amount of charge through a circuit in a certain time interval as is sent by an alternating current through the same circuit in the same time interval.To calculate its value, let an alternating current be represented byI=I0sin𝜔tThe charge sent by the alternating current I in time dt is given bya

	dq=Idt[∴I=dq dt]
	dq=I0sin𝜔tdt

The amount of charge passing through the circuit in half time period can be obtained by integrating above equation (i) from t=0 to t=T/2.a

q	=T/2∫0I0sin𝜔tdt=I0T/2∫0 sin wtdt
	=I0[-cos𝜔t 𝜔]T/2 0
	=-I0 𝜔[cos𝜔t]T/20
	=-I0 2𝜋/T[cos2𝜋 Tt]T/20[∴w=2𝜋 T]
	=-I0T 2𝜋[cos2𝜋 TT 2-cos0]
	=-I0T 2𝜋[cos𝜋-cos0]
	=-I0T 2𝜋[-1-1]
	=-I0T 2𝜋×-2
q	=I0T 𝜋…( ii )

If Im be the mean value of an a.c. over positive half cycle, then the charge sent by it in time T/2 is given byq=InT

2From equation (ii) and (iii), we geta

ImT 2	=I0T 𝜋
∴Im	=2I0 𝜋
	=0.637I0

Hence it means an average value of a.c. cower positive half cycle is 0.637 times the peak value of a.c. is 63.7% of the peak value. Similarly, the mean or average value of a.c. over the negative half cycle is obtained by integrating equation (1) from t=T/2 to t=T. It comes out to be 0.637Io. Hence the mean or average value of a.c aver one complete cycle is 0.637I0-0.637I0= zeroSimilarly, the mean value of alternating e.m.f isEm=2E0

𝜋=0.657E0Phasor Diagram

A phasor is a vector rotating at a constant angular velocity. The sinusoidal output voltage produced by the simple generator can be represented by the phasor diagram shown in Figure. A diagram containing these phasors is called phase diagram. Science an alternating current or emf varies sinusoidally, it can be represented in a phase diagram.Wave diagram:It is the displacements of current and voltage in a circuit are represented in a graph by their wave forms along with their phase difference, the diagram is called wave diagram. Let an alternate current,I = Iosin 𝜔twhere, 𝜔= 2𝜋f, ( f being frequency)