AC Fundamental. Simple Loop Generator: Whenever a conductor moves in a magnetic field, an emf is induced in it.

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1 A Fundaental Siple oop Generator: Whenever a conductor oves in a agnetic field, an ef is induced in it. Fig.: Siple oop Generator The aount of EMF induced into a coil cutting the agnetic lines of force is deterined by the following three factors.. Speed the speed at which the coil rotates inside the agnetic field.. Strength the strength of the agnetic field. 3. ength the length of the coil or conductor passing through the agnetic field. Prepared by: Nafees Ahed

"nstantaneous value = Maxiu value x sinθ and this is generalized by the forula.

2 What is instantaneous value? Fig.: nstantaneous alue The instantaneous values of a sinusoidal wavefor is given as the "nstantaneous value = Maxiu value x sinθ and this is generalized by the forula. v Sin Sinusoidal Wavefor onstruction oil Angle ( θ ) e = ax.sinθ Fig.3: Wave for construction Prepared by: Nafees Ahed

3 Sinusoidal Quantities (oltage & urrent) oltage or EMF is denoted by v Sint O e E Sint Fig.4 Where urrent or E = Maxiu alue of oltage or EMF v or e = nstantaneous value of oltage or EMF f Angular frequency f = frequency in Hz i Sint Note: n A electrical theory every power source supplies a voltage that is either a sine wave of one particular frequency or can be considered as a su of sine waves of differing frequencies. The good thing about a sine wave such as (t) = Asin(ωt + δ) is that it can be considered to be directly related to a vector of length A revolving in a circle with angular velocity ω. The phase constant δ is the starting angle at t =. Following aniated GF link shows this relation. sine.gif Phase difference, Phasor diagra and eading & lagging concepts: When two sine waves are produced on the sae display, they ay have soe phase different, one wave is often said to be leading or lagging the other. onsider following two sine waves v=sin(ωt+45) v=sin(ωt+3) 45 3 Fig.5 Phasor Diagra (lockwise +e angle) 3 Prepared by: Nafees Ahed

4 Here v & v are having a phase difference of 5.The blue(v) vector is said to be leading the red(v) vector Or onversely the red vector is lagging the blue vector. This terinology akes sense in the revolving vector picture as shown in following GF figure. doublesine.gif Displaced wavefors: Fig. 6 Soe Terinology. Wave for: The shape of the curve.. nstantaneous value: The value at any instant of tie. 3. ycle: One coplete set of +ve and ve values 4. Tie Period: Tie taken to coplete one cycle. 5. Frequency: Nuber of cycles copleted in one second. f Hz T Average value of A et i Sint Average current Area of half cycle av idt Sint dt Fig Prepared by: Nafees Ahed

5 Sint dt cost Note: Average value of the following for coplete cycle = av sin t cos t Siilarly average value of ac voltage sin( t ) av cos( t ) MS (oot Mean Square or Effective) alue: et i Sint MS current rs Area of cycle of i dt i Sin t dt cost t sin t d t 4 rs Siilarly MS value of ac voltage rs Fig.8 Peak Factor(Kp): Maxiu alue K p.44 for coplete sine wave MS alue / For Factor (Kf): MS alue / K f. for coplete sine wave Avarage alue / Exaple : alculate the average current, effective voltage, peak factor & for factor of the output wavefor of the half wave rectifier. Solution: 5 Prepared by: Nafees Ahed

6 Solution: Fig.9 avg rs Sint dt cos sin Sin t dt t sin t 4 K K p f Maxiu alue MS alue / MS alue Avarage alue /.57 / Question: Find the above for the output wavefor of the full wave rectifier. : avg, rs, K p. 44 & K f. Addition and subtraction of alternating quantities: onsider the following exaples Exaple : Three sinusoidal voltages acting in series are given by v= Sin44t v= Sin(44t-45) v3= os44t Deterine (i) an expression for resultant voltage (ii) the frequency and rs value of resultant voltage. Solution: Phasor diagra ethod: oltage v3 ay be rewritten as v3= Sin(44t+9 ) Phasor diagra all voltages will be as shown bellow (Taking axiu values) Y axis 3= 6 = axis 45 Fig. : Phasor Diagra = Prepared by: Nafees Ahed

7 (i) coponent of axiu value of resultant voltage x= os + os45 + os9 = Y coponent of axiu value of resultant voltage y= Sin - Sin45 + Sin9 = So axiu value of resultant voltage x y y tan x So resultant voltage v= Sin(ωt+ɸ) v= 5 Sin(44t+6.56 ) (ii) ω=44 or ᴫf=44 => f=7 Hz rs=/ =5.8 Question: Draw the phasor diagra of following wavefor v= Sin5t v= Sin(5t+π/3) v3=-5 os5t v4=5 Sin(5t-π/4) Note: v3=-5 os5t=+5 Sin(5t-π/) 5 Single phase A circuit:. Purely esistive ircuit: et applied voltage v = Sinωt () According to Oh s aw v=i Sinωt = i i =/ Sinωt et / = So i = Sinωt () Phasor diagra v ~ Fig. a Fig.b Φ =, PF cos Φ=, No phase different or & are in sae phase Note:. osine of angle between oltage () and current () is known as power factor i.e PF= osɸ. f current lags => power factor lags 3. f current leads => power factor leads 7 Prepared by: Nafees Ahed

8 Power consued nstantaneous power p=v*i = Sinωt* Sinωt = Sin ωt Average power P Wavefor pdt Sin t dt cost P dt P rs rs dt cost Fig. c Note: t is also clear fro the above graph that p is having soe average value. Exaple 3: A 5 olts (MS), 5 Hz voltage is applied across a circuit consisting of a pure resistance of oh. Deterine i. The current flowing through the circuit ii. Power absorbed by circuit iii. Expression for voltage and current iv. Draw wavefor and phasor diagra. Solution: i. 5 rs. 5Ap ii. Power absorbed ==5*.5=35 Watt iii. v 5 sin (5) t 5 sin t i.5sin t iv. See figure c 8 Prepared by: Nafees Ahed

9 . Purely nductive ircuit: et applied voltage v = Sinωt () Fig. a ~ v When an alternating voltage is applied to purely inductive coil, an ef known as self induced ef is induced in the coil which opposes the applied voltage. Self induced ef di( t) e dt Since applied voltage at every instant is equal and opposite to self induced ef We have v=-e di( t) Sin t dt di Sint dt ntegrating ost i Sint dt / Sin t et Here ω= know as inductive reactance So i Sin t / Here current lags by 9. Phasor diagra Φ =9, PF osφ=, Phase different between & is 9. urrent () lags by 9 fro voltage (). Power consued nstantaneous power p=v*i = Sinωt* Sin(ωt-9 ) = - Sinωt osωt = - / Sinωt Average power pdt P Ava Sint Fig.b Hence NO power is consued in purely inductive circuit. 9 Prepared by: Nafees Ahed

10 Wavefor Fig. c Note: t is also clear fro the above graph that p is having EO average value. Exaple 4: The voltage and current thorough a circuit eleent are v sin(34t 45 ) and i sin(34 t 35 ) A. i. dentify the circuit eleents ii. Find the value iii. Obtain expression for power Solution: i. Phase difference = =9 and lags v by 9, see the phasor diagra. So it is pure inductor Fig.3 ii. nductance =ω => =34 => =/34 H iii. p sin(34t 45 ) sin(34t 35 ) sin 68t 3. Purely apacitive ircuit: et applied voltage v = Sinωt () Fig. 4a ~ v Prepared by: Nafees Ahed

11 harge And we know et Here Q v Q Sint i dq dt d dt Sint dt ost Sint / i So Sint / i Here current leads by 9. know as capacitive reactance Phasor diagra Φ =9, PF osφ=, Phase different between & is 9. urrent () leads by 9 fro voltage (). Fig.4b Power consued nstantaneous power p=v*i = Sinωt* Sin(ωt+9 ) = Sinωt osωt = / Sinωt Average power pdt P Ava Sint Hence NO power is consued in purely capacitive circuit. Wavefor Fig. 4c Prepared by: Nafees Ahed

12 Note: t is also clear fro the above graph that p is having EO average value. Exaple 5: A 38 μf capacitor is connected to, 5 Hz supply. Deterine i. apacitive reactance offered by the capacitor ii. The axiu current iii. MS value of current drawn by capacitor iv. Expression for voltage & current Solution: i. =/ω=/ᴫf= Oh ii. =/= /= A iii. rs = / = A iv. v sin t i sin t / 4. - ircuit: and et the applied voltage is oltage across resistance oltage across inductance Fro phasor diagra v = Sinωt () =. =. =.(ω) =. (πf) ( ) ( Where know as ipedance Fro phasor diagra, is lagging behind by an angle ϕ tan tan So Sint Where i ) Fig. 5a Phasor diagra Phase different between & is ϕ. urrent () lags by ϕ fro voltage (). & are in sae ϕ phase, is leading by an angle of 9 fro etc. Fig.5b Prepared by: Nafees Ahed

13 Power consued nstantaneous power p=v*i = Sinωt* Sin(ωt-ϕ ) SintSin t cos cos t Average power P Ava cos cos t Ava cos Ava cos t P cos Ava cost Wavefor P cos cos Fig. 5c oltage triangle and pedance triangle: Dividing each side by ϕ ϕ oltage Triangle Power factor: os ϕ=/=/ Fig. 5d pedance Triangle 3 Prepared by: Nafees Ahed

14 Active power, reactive power, apparent power and power factor: Active power P = os ϕ = Unit: Watt, KW eactive power Q = Sin ϕ = Unit: A, KA Apparent Power S = = Unit: A, KA S Q ϕ P Power Triangle Fig. 5e Power factor: cos ϕ= Active power/apparent power Exaple 6: A resistance and inductance are connected in series across a voltage v 83 sin(34 t) & current expression as i 4sin(34t / 4). Find the value of resistance, inductance and power factor. Solution: =/=7.75 = cosϕ=7.75* cos(ᴫ/4)=5.7 oh = sinϕ=7.75* sin(ᴫ/4)=5.7 oh =ω => =.59 H Power factor = cos ϕ= cos(ᴫ/4)=/ =.77 Exaple 7: i 4.4 sin( t / 6) passes in an electric circuit when a voltage of v 4.4sin t is applied to it. Deterine the power factor of the circuit, the value of true power, apparent power and circuit coponents. Solution: =/=4.4/4.4= Oh PF cosϕ=cos(ᴫ/6)=.866 P= cosϕ =(4.4/ )( 4.4/ )*.866=866 W S==(4.4/ )( 4.4/ )= W =cosϕ=*.866=8.66 =sinϕ=*.5=5 Oh 5. - ircuit: et the applied voltage is oltage across resistance oltage across inductance Fro phasor diagra v = Sinωt () =. Fig. 6a =. =.(/ω) =. (/πf) 4 Prepared by: Nafees Ahed

15 ( ) ( ) Where know as ipedance Fro phasor diagra, is leading by an angle ϕ tan tan So Sint Where i Phasor diagra Phase different between & is ϕ. urrent () leads by ϕ fro voltage (). & are in sae phase, is lagging by an angle of 9 fro etc. ϕ Fig.6b Power consued nstantaneous power p=v*i = Sinωt* Sin(ωt+ϕ ) sin tsin t cos cos t Siilar to - circuit average power P cos Wavefor 5 Fig. 6c Prepared by: Nafees Ahed

16 Note: Siilar to - circuit we can have voltage triangle and power triangle ircuit: Fig. 7a et the applied voltage is v = Sinωt () oltage across resistance oltage across inductance oltage across inductance =. =. =. Fro phasor diagra ) ( ) Where ( know as ipedance Fro phasor diagra, is lagging by an angle ϕ tan So Sin t Where i tan Phasor diagra - ϕ 6 Fig.7b Assuing > Prepared by: Nafees Ahed

17 Power consued Average power (siilar to - circuit) P cos Different cases:. > => nductive circuit, lags. < => apacitive circuit, leads 3. = => esistive circuit, & are in sae phase, power factor PF is Unity. This condition is called as electrical resonance. Exaple 8: A series circuit consisting of resistance of Oh, inductance of. H and capacitance of 5 μf is connected across a 3, 5 Hz source. alculate. urrent. Magnitude & nature of power factor Solution: = Oh =ᴫf=ᴫ*5*.=6.8 Oh =/ᴫf=/(ᴫ*5*5* -6 )= i. =/=4.98 A ii. PF=/=.433 lagging (As >) pedance(): Where = esistance Ω = eactance Ω PF: osϕ= / Ω, Oh ϕ pedance Triangle Fig. 8 Adittance(Y): t is reciprocal of ipedance Y=/ Y Where G= onductance Ω - = Susceptance Ω - PF: osϕ= G/Y G=Yos ϕ=(/)(/)=/. B=YSin ϕ=(/)(/)=/. G B Ω -, Mho, Sieens Y ϕ G B AdittanceTriangle Fig Prepared by: Nafees Ahed

18 Parallel ircuits: There are 3 ethods for solving parallel circuits. Phasor Method: onsider the following circuit For branch cos (eading PF) if > (nductive circuit) (agging PF) if < (apacitive circuit) Siilarly for branch and 3 we have,, osɸ and 3, 3, osɸ3 respectively. Now we can draw the phasor diagra as shown bellow assuing any values for,, 3 and ɸ, ɸ, ɸ3 + - Fig. a 3 Φ3 Φ Φ Fig.b esultant current which will be the phasor su of, & 3 can be deterine by using any of following ethod i. Parallelogra ethod ii. esolving branch current, & 3 along x and y-axis. oponent of resultant current along x-axis os os os 3os 3 oponent of resultant current along x-axis Sin Sin Sin Sin Prepared by: Nafees Ahed

19 Sin os & tan. Adittance Method: onsider the following circuit Sin os, Y, G, B, Y, G, B 3, Y, G, B Fig. We know conductance G => find out G, G & G3 We know susceptance B => find out B, B & B3 Total conductance G=G+G+G3 Total susceptance B=B+B+B3 Total Adittance Y G Y So = => =/=Y =Y =Y 3=Y3 3. Sybolic or j-notation Method: onsider following voltage and current v sin t i sin t Wave for and phasor diagra will be as shown bellow 9 Prepared by: Nafees Ahed

20 Fig. a Therefore either the tie wavefor of the rotating phasor or the phasor diagra can be used to describe the syste. Since both the diagras, the tie diagra and the phasor diagra convey the sae inforation, the phasor diagra being uch ore sipler, it is used for an explanation in circuit theory analysis. Since electrical data is given in ters of rs value, we draw phasor diagra with phasor values as rs rather than peak value used so far. Above voltage can be represented by = x + jy artesian o-ordinates = θ in polar coordinates. Here x=oponent of along x-axis y= oponent of along y-axis j is an operator which when ultiplied to a phasor rotates the phasor 9 anticlockwise and j = ( ) or j = - Fig. b: Phasor epresentation Hence a voltage or current can be represented by a coplex nuber. Note: n Phasor algebra: o Addition & Subtraction is done in artesian for. o Multiplication, Division, power and roots are done in polar for. Prepared by: Nafees Ahed

21 Phasor Algebra applied to single phase circuit. - series circuit: Fig. 3 onsider as reference j oltage drop across resistance =+j oltage drop across inductance =+j So total voltage =+ =( +j)+( +j) =+j =(+j) = Where =+j= onsider as reference =+j=. - series circuit: Fig. 4 onsider as reference j oltage drop across resistance =+j oltage drop across capatance =-j So total voltage =+ =( +j)+( -j) =-j =(-j) = Where =-j= onsider as reference =+j= Prepared by: Nafees Ahed

22 Power deterination: et v=+3j volts & i=3-j Ap v i ( 3 j) (3 j) 3 j Here e v i Active power or real power = =3 Watt eactive power = v i = A. -- parallel circuit: For branch : =+j Where = j j j j j j j Siilarly for branch & 3 j Where = j Where 3 = So total current =++3 Series-parallel circuit: + - Fig. 5 pedances in series eq=++3+. pedances in parallel Prepared by: Nafees Ahed

23 eq 3... A series parallel ac circuit can be solved in sae anner as that of D series parallel circuit except that coplex ipedances are used instead of resistances. Exaple 9: A 5H inductor is in series with a parallel cobination of a oh resistance & μf capacitor. f ω of applied voltage is. Find i. Total ipedance ii. Total adittance iii. urrent in each branch if applied voltage is 3 Solution: =5H =8Ω =μf Fig. 6 ==8 Ω =-j= 6 j 3=j=jω=jx5x -3 Ω =5jΩ Equivalent ipedance 8 5 j j j = j = i. Y j ii j iii. By current division rule 5 j j j By current division rule O =-= ( j )-( j )= j 3 Prepared by: Nafees Ahed

AC Fundamental. Simple Loop Generator: Whenever a conductor moves in a magnetic field, an emf is induced in it.

AC Fundamental. Simple Loop Generator: Whenever a conductor moves in a magnetic field, an emf is induced in it. AC Fundamental Simple Loop Generator: Whenever a conductor moves in a magnetic field, an emf is induced in it. Fig.: Simple Loop Generator The amount of EMF induced into a coil cutting the magnetic lines