Alternating current

An alternating voltage or current is one whose magnitude changes with time and direction reverse periodically.

The mean value of the ac is defined as equivalent steady current which when passed through a circuit would send the same amount of charge in the same time as is done by the alternating current in the same circuit in the same time. RMS value of an alternating current is defined as that steady current which when passed through a given time produces the same amount of heat as produced by alternating current when passes through the same resistance for the same time.

Peak value and r.m.s value:

Peak value of the ac is defined as a maximum value or the amplitude of an ac .The root mean square (rms) value of AC is equivalent to that steady current which when passed through a resistor would produce the same heat that would be done by the AC in passing through the same resistor in similar condition(same time).

To calculate the mean value of ac, Let us consider an ac given by

I= I0sinωt………..1

The small amount of work done by an ac in small time dt is

dW=I2Rdt……………….2

where R is the resistance

So total work done is obtained by integrating equation ii from t = 0 to t = T we get

W=I20R∫0Tsin2ωt${\mathrm{I}}_0^2\mathrm{R}\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{s}\mathrm{i}{\mathrm{n}}^2\omega \mathrm{t}$

Or, W=I20R∫0T1−cos2ωt2dt${\mathrm{I}}_0^2\mathrm{R}\underset{0}{\overset{\mathrm{T}}{\int }}\frac{1-\mathrm{c}\mathrm{o}\mathrm{s}2\omega \mathrm{t}}{2}\mathrm{d}\mathrm{t}$

Or, W=I20R2∫0Tdt$=\frac{{\mathrm{I}}_0^2\mathrm{R}}{2}\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{d}\mathrm{t}$

Or, W=I20R2{∫0Tdt−∫0Tcos2ωtdt}$=\frac{{\mathrm{I}}_0^2\mathrm{R}}{2}\{\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{d}\mathrm{t}-\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{c}\mathrm{o}\mathrm{s}2\omega \mathrm{t}\mathrm{d}\mathrm{t}\}$

 Or, W=I20R2T$\frac{{\mathrm{I}}_0^2\mathrm{R}}{2}\mathrm{T}$………..3

If Irms be the RMS value of a.c. then it would do the same work as in equation iii while passing through R in time T

W=Irms2RT……………4

Now from equation iii and iv we get

Irms2RT=I20R2T$\frac{{\mathrm{I}}_0^2\mathrm{R}}{2}\mathrm{T}$

Or, Irms.=I0.12√$\frac{1}{\sqrt{2}}$

Or, Irms.=0.707I0

This relation shows that the root mean square value of ac is about 70.7% of its pick value.

Similarly the root mean square value of alternating emf is

Erms=0.707E0

Phasor diagram:

A vector diagram in which alternating voltage or current are treated as vectors and represented by straight lines with arrows and difference of phase angle between them are called phasor diagram.

The straight line representing the current while rotates in the phase diagram with the same angular frequency (w) as the displacement of the current in the AC current is called phasor.

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.

Relation between current and voltage when an alternating current passes:

The straight line representing the current while rotates in the phase diagram with the same angular frequency (w) as the displacement of the current in the AC current is called phasor.

Consider a pure inductor of inductance L connected to alternating source E having frequency ‘f’ as shown in figure.

Let instantaneous value of alternating current is:

I = Iosinwt…(i)

The potential difference across the inductor,

V = −LdIdt$\frac{-\mathrm{L}\mathrm{d}\mathrm{I}}{\mathrm{d}\mathrm{t}}$

Using Krichoff’s law,

E + V = 0

Or, E – L.dIdt$\frac{\mathrm{d}\mathrm{I}}{\mathrm{d}\mathrm{t}}$ = 0

Or, E = L.dIdt$\frac{\mathrm{d}\mathrm{I}}{\mathrm{d}\mathrm{t}}$

Or, E = LdIosinwtdt$\frac{\mathrm{L}\mathrm{d}{\mathrm{I}}_{\mathrm{o}}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{w}\mathrm{t}}{\mathrm{d}\mathrm{t}}$

Or, E = LIo(wr.wt).w

Or, E = LW.IOwrw2

Or, E = XLIocoswt.

Where, LW is called inductive reactance. It is denoted by XL.

Or, E = Eocoswt.

Where, (XLIo) = Eo is called maximum value of emf.

E = Eosin(wt + π2$\frac{\pi }{2}$)…(ii)

From equation (i) and (ii) it follows that the alternating emf leads the current by phase angle π2$\frac{\pi }{2}$.

Expression for the impedance of the circuit:

Let us consider C is the capacitance of the capacitor and R is the resistance of the resistor connected in the series with a.c source having e.m.f E.with frequency f. and the voltage across the capacitor is VC and voltage across the resistance is VR.

In case of resistor, voltage (VR) and current (I) both are in same phase but is case of capacitor, voltage (VC) lags the current (I) by ∏/2 . Then from the phasor diagram we have

E2=V2R+V2C${\mathrm{E}}^2={\mathrm{V}}_{\mathrm{R}}^2+{\mathrm{V}}_{\mathrm{C}}^2$

Or, E2=I2(R2+X2C)${\mathrm{E}}^2={\mathrm{I}}^2({\mathrm{R}}_{}^2+{\mathrm{X}}_{\mathrm{C}}^2)$ where VR=IR and VC=I XC

Z=E/I=R2+X2C−−−−−−−√$\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{C}}^2}$ where Z=E/I called the impedence of the RC circuit.

Also we have

I=E(R2+X2C)√=$\frac{\mathrm{E}}{\sqrt{({\mathrm{R}}_{}^2+{\mathrm{X}}_{\mathrm{C}}^2)}}=$E(R2+(1cω)2)√$\frac{\mathrm{E}}{\sqrt{({\mathrm{R}}_{}^2+{\left(\frac{1}{\mathrm{c}\omega }\right)}_{}^2)}}$

Suppose ɵ be the angle through which the current lage on the voltage

Then, from phase diagram we have tanɵ=−VCVR$-\frac{{\mathrm{V}}_{\mathrm{C}}}{{\mathrm{V}}_{\mathrm{R}}}$ =−XLR$\frac{-{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}$

Here negative sign indicate that the voltage lag behind the current which is same as I lead E

Phase relation between current and voltage:

Suppose L be the inductance of the inductor, C be the capacitance of a capacitor and R be the resistance are connected in the series with a.c source having emf.E and frequency f .

And the voltage across the inductor is VL, the voltage across the capacitor is VCand the voltage across the resistance.

Figure.

Now the p.d across the inductor having resistance XL

VL=IXL ( in this case VL lead I by ∏/2)

p.d across the capacitor having the resistance XC is

VC=IXC ( in this case VC lags the current I by ∏/2)

p.d across the capacitor having the resistance R is

VR=IR ( in this case VR and I are in same phase )

Since the inductor, resistor and capacitor are connected in series then the current in all resistance are same but voltage depend upon the resistance i.e. if XL>XC then VL> VC. Then from phasor diagram:

We have

(OB)2=(OA)2+(BA)2  (by vector sum)

E2=V2R${\mathrm{V}}_{\mathrm{R}}^2$+(VL−VC)2${\left({\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{C}}\right)}^2$       from phasor diagram BA=OD= VL-VC

Or,E=IR2+(XL−XC)2−−−−−−−−−−−−−−√$\mathrm{I}\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$ from above

Or, Z=E/I =R2+(XL−XC)2−−−−−−−−−−−−−−√$\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$

Where Z=E/I is the impedance of LRC circuit

Phase relation:

From figure we have

tan⁡ɵ=BAOP=(VL−VC)/VR=(XL−XC)R$\frac{\mathrm{B}\mathrm{A}}{\mathrm{O}\mathrm{P}}=\left({\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{C}}\right)/{\mathrm{V}}_{\mathrm{R}}=\frac{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}{\mathrm{R}}$

ɵ

Therefore XL is greater then that XC so, tanɵ is positive this show the E lead I.

But if XC>XL then we have

ɵ

thisshowthat$\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{h}\mathrm{o}\mathrm{w}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}$Then tanɵ is negative which show that the E lags I

If θ = 0 I and E are in phase.

The impedance of an LCR series circuit:

Suppose L be the inductance of the inductor, C be the capacitance of a capacitor and R be the resistance are connected in the series with a.c source having emf. E and frequency f .

And the voltage across the inductor is VL, the voltage across the capacitor is VC and the voltage across the resistance.

Now the p.d across the inductor having resistance XL

VL=IXL ( in this case VL lead I by ∏/2)

p.d across the capacitor having the resistance XC is

VC=IXC ( in this case VC lags the current I by ∏/2)

p.d across the capacitor having the resistance R is

VR=IR ( in this case VR and I are in same phase )

Since the inductor, resistor and capacitor are connected in series then the current in all resistance are same but voltage depend upon the resistance i.e. if XL>XC then VL> VC. then from phasor diagram we have

(OB)2=(OA)2+(BA)2  (by vector sum)

E2=V2R${\mathrm{V}}_{\mathrm{R}}^2$+(VL−VC)2${\left({\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{C}}\right)}^2$       from phasor diagram BA=OD= VL-VC

Or,E=IR2+(XL−XC)2−−−−−−−−−−−−−−√$\mathrm{I}\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$ from above

Or, Z=E/I =R2+(XL−XC)2−−−−−−−−−−−−−−√$\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$

Where Z=E/I is the impedance of LRC circuit

Phase relation:

From figure we have

Tan ɵ

ɵ

Therefore XL is grater then that XC so, tanɵ is positive this show the E leads I.

But if XC>XL then we have

ɵ

thisshowthat$\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{h}\mathrm{o}\mathrm{w}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}$Then tanɵ is negative which show that the E lags I

At the resonance the inductance reactance(X L) is equal to the capacitive reactance XC i.e.

XL=XC

Lω=1/Cω

Or, ω2=1/LC

Or, f=12πLC√$\frac{1}{2\pi \sqrt{\mathrm{L}\mathrm{C}}}$       where ω= 2π f

Here fis the resonant frequency. The resonant frequency is independent of the resistance in the circuit from the above result. But the sharpness of the resonance decrease with increase in the resistance. it is used to filter the frequency and also used in the radios, television and telephone carried equipments.

Expression for AC power:

Let the alternating E.M.F. applied t LCR circuit be:

E = Eosinwt ..(i)

If the alternating current developed lags behind the applied emf by a phase angle θ, then,

I = Iosin(wt – v)…(ii)

Power at instant t is given by:

Or, dwdt$\frac{\mathrm{d}\mathrm{w}}{\mathrm{d}\mathrm{t}}$ = EI

= Eosinwt * Iosin(wt – θ)

= EoIosinwt(sinwtcosθ – coswtsinθ)

= EoIosin2wtcosθ – EoIosinwt.coswt.sinθ

= EoIosin2wt.cosθ – EoIo2$\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}$sin2wt.sinθ.

If this instantaneous power is assumed to remain constant for a small time dt, then small amount of work done in this time is

Or, dw = [EoIosin2wt.cosθ – EoIo2$\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}$sin2wtcosθ]dt

Total work done over a complete cycle is obtained by integrating above equation from 0 to T.

Ie. w = ∫0TEoIosin2wt.cosθdt−∫0TEoIo2sin2wt.sinθdt$\underset{0}{\overset{\mathrm{T}}{\int }}{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}{\sin }^2\mathrm{w}\mathrm{t}.\mathrm{c}\mathrm{o}\mathrm{s}\theta \mathrm{d}\mathrm{t}-\underset{0}{\overset{\mathrm{T}}{\int }}\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{w}\mathrm{t}.\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{d}\mathrm{t}$…..(iii)

Or, W = EoIocosθ∫0Tsin2wt.dt−EoIo2sinθ∫0Tsin2wt.dt$\underset{0}{\overset{\mathrm{T}}{\int }}{\sin }^2\mathrm{w}\mathrm{t}.\mathrm{d}\mathrm{t}-\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}}{2}\mathrm{s}\mathrm{i}\mathrm{n}\theta \underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{w}\mathrm{t}.\mathrm{d}\mathrm{t}$

Now,

Or, ∫0Tsin2wt.dt=∫0T(1−cos2wt2)dt=12[∫0Tdt−∫0Tw.2wt.dt]$\underset{0}{\overset{\mathrm{T}}{\int }}{\sin }^2\mathrm{w}\mathrm{t}.\mathrm{d}\mathrm{t}=\underset{0}{\overset{\mathrm{T}}{\int }}\left(\frac{1-\mathrm{c}\mathrm{o}\mathrm{s}2\mathrm{w}\mathrm{t}}{2}\right)\mathrm{d}\mathrm{t}=\frac{1}{2}\left[\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{d}\mathrm{t}-\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{w}.2\mathrm{w}\mathrm{t}.\mathrm{d}\mathrm{t}\right]$

= 12[T−0]$\frac{1}{2}\left[\mathrm{T}-0\right]$ = T2$\frac{\mathrm{T}}{2}$….(iv)

And, ∫0Tsin2wt.dt$\underset{0}{\overset{\mathrm{T}}{\int }}\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{w}\mathrm{t}.\mathrm{d}\mathrm{t}$ = 0 ….(iv)

Using equation (iv) and (v) in equation (iii), we get,

Or, W = EoIowsθT2$\frac{\mathrm{T}}{2}$ – 0

Or, W = EoIowsθ * TZ$\frac{\mathrm{T}}{\mathrm{Z}}$

So, Average power in the inductive circuit over a complete cycle,

Or, P = wT$\frac{\mathrm{w}}{\mathrm{T}}$ = EoIocosθT.T2$\frac{{\mathrm{E}}_{\mathrm{o}}{\mathrm{I}}_{\mathrm{o}}\mathrm{c}\mathrm{o}\mathrm{s}\theta }{\mathrm{T}}.\frac{\mathrm{T}}{2}$ = Eo2√.Io2√.cosθ$\frac{{\mathrm{E}}_{\mathrm{o}}}{\sqrt{2}}.\frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}.\mathrm{c}\mathrm{o}\mathrm{s}\theta$

So, P = EVIVcosθ….(v)

Here, O is called true power and EVIV is called apparent or virtual power.

Ratio of true power and apparent power in an A.C. circuit is called power factor of the circuit.

Choke coil:

A choke coil is a coil of a number of turns and negligible resistance wounded on a soft iron laminated core. Such a coil has induced L and resistance zero. The power consumed in the coil is given by:

P = Irms* Erms * Power Factor.

Where, power factor = cosθ = RR2+L2W2√$\frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{L}}^2{\mathrm{W}}^2}}$.

Since, for an ideal coil, R = 0, the power consumed by a coil P = 0 watt. This implies that a coil having zero ohmic resistance consumes no power from the source. However, the coil offers resistance to the alternating current as it is given by XL = wL = 2πfL. It is due to these reasons; such a coil is used for limiting as current. Hence, choke will be preferred to a resistance in AC circuit.

Electrical resonance:

When the frequency of the applied emf is equal to natural frequency in the electrical circuit the current reaches to the maximum value and the circuit is said to be in the electrical resonance.

Telecommunication:

Frequency at the resonance is given by, f =12πLC√$\frac{1}{2\pi \sqrt{\mathrm{L}\mathrm{C}}}$    series resonance circuit has great application in telecommunication.  A radio tuner circuit uses the series resonant circuit to tune the radio the capacitor of the circuit is varied in such way that the frequency of circuit matches with frequency of the station to be received.Space consists of the various frequencies corresponding to the various TV stations and radio. The series resonant circuit only accept the signal through the antenna which tune the series resonant circuit .the  accepted frequency has very high amplitude current in the circuit and amplified by the electronic circuit .In this way series resonant circuit is used in telecommunication.

An inductor having self inductance L is used in a ac power supply:

It is the property of the electrical circuit which opposes the strength of the current flowing through the circuit by inducing an emf or current in itself is called self inductance.

Let L be the inductor of the inductance and I be the current pass through them at any instant t. then from faraday’s law induction we have

E=-dɸ/dt

Since ɸ=LI for self inductance

Then E= -d(LI)/dt……….1

If E be the emf anf I be the current then we have power P=EI…………..2

From 1 and 2 we have

P=-I.d(LI)/dt…………3

Now the small wok done from eq. 3 we have dW = Pdt=LI2……….4

Taking the integration from 0 to W and 0 to I of equation 4 we get

W=12LI2$\frac{1}{2}\mathrm{L}{\mathrm{I}}^2$

This is the equation which so that the energy store in the magnetic field.