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

Transcription

1 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 of force is determined by the following three factors.. Speed the speed at which the coil rotates inside the magnetic field.. Strength the strength of the magnetic field. 3. Length the length of the coil or conductor passing through the magnetic field.

2 What is instantaneous value? Fig.: nstantaneous alue The instantaneous values of a sinusoidal waveform is given as the "nstantaneous value = Maximum value x sinθ and this is generalized by the formula. v m Sin Sinusoidal Waveform Construction Coil Angle ( θ ) e = max.sinθ Fig.3: Wave form construction

3 Sinusoidal Quantities (oltage & Current) v oltage or EMF is denoted by v m Sin t O e E m Sin t m p p?t Fig.4 Where m or Em = Maximum alue of oltage or EMF v or e = nstantaneous value of oltage or EMF f Angular frequency f = frequency in Hz Current i m Sin t Note: n AC electrical theory every power source supplies a voltage that is either a sine wave of one particular frequency or can be considered as a sum 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 animated GF link shows this relation. sine.gif Phase difference, Phasor diagram and Leading & lagging concepts: When two sine waves are produced on the same display, they may have some phase different, one wave is often said to be leading or lagging the other. Consider following two sine waves v=msin(ωt+45) v=msin(ωt+3) 45 3 Phasor Diagram (Clockwise +e angle) Fig.5 3

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 Conversely the red vector is lagging the blue vector. This terminology makes sense in the revolving vector picture as shown in following GF figure. doublesine.gif Displaced waveforms: Fig. 6 Some Terminology Wave form: nstantaneous value: Cycle: Time Period: Frequency: The shape of the curve. The value at any instant of time. One complete set of +ve and ve values Time taken to complete one cycle. Number of cycles completed in one second. f Hz T Average value of AC Let i m Sin t Average current Area of half cycle av 4 id t m i m p p?t Fig.7 Sin t d t

5 m Sin t d t m cos t Note: Average value of the following for complete cycle = sin t cos t sin( t ) cos( t ) m Similarly average value of ac voltage av m MS (oot Mean Square or Effective) alue: av Let i m Sin t MS current i Area of cycle of i rms i d t i π m Sin t d t π ωt m cos t d t m t sin t 4 Fig.8 m Similarly MS value of ac voltage rms m rms Peak Factor(Kp): Kp m Maximum alue.44 MS alue m / for complete sine wave Form Factor (Kf): Kf / MS alue m. Avarage alue m / for complete sine wave Example : Calculate the average current, effective voltage, peak factor & form factor of the output waveform of the half wave rectifier. Solution: 5

6 v m π π Solution: Fig.9 m avg Sin t d t rms ωt 3π m m cos sin m Sin t d t m t sin t m 4 Maximum alue m MS alue m / / MS alue Kf m.57 Avarage alue m / Kp Question: Find the above for the output waveform of the full wave rectifier. : avg m, rms m, K p.44 & K f. Addition and subtraction of alternating quantities: Consider the following examples Example : Three sinusoidal voltages acting in series are given by v= Sin44t v= Sin(44t-45) v3= Cos44t Determine (i) an expression for resultant voltage (ii) the frequency and rms value of resultant voltage. Solution: Phasor diagram method: oltage v3 may be rewritten as v3= Sin(44t+9) Phasor diagram all voltages will be as shown bellow (Taking maximum values) Y axis m3= m= 45 6 X axis m= Fig. : Phasor Diagram

7 (i) X component of maximum value of resultant voltage mx= Cos+ Cos45+ Cos9= Y component of maximum value of resultant voltage my= Sin - Sin45 + Sin9 = So maximum value of resultant voltage m mx my 4 5 tan my 6.56 mx So resultant voltage v=m Sin(ωt+ɸ) v= 5 Sin(44t+6.56) ω=44 or ᴫf=44 => f=7 Hz rms=m/ =5.8 (ii) Question: Draw the phasor diagram of following waveform v= Sin5t v= Sin(5t+π/3) v3=-5 Cos5t v4=5 Sin(5t-π/4) Note: v3=-5 Cos5t=+5 Sin(5t-π/) Single phase AC circuit:. Purely esistive Circuit: Let applied voltage v =m Sinωt () According to Ohm s Law v=i m Sinωt = i i =m/ Sinωt Let m/ =m So i =m Sinωt () Phasor diagram ~ v Fig. a Fig.b Φ =, PF cos Φ=, No phase different or & are in same phase Note:. Cosine of angle between oltage () and current () is known as power factor i.e PF= Cosɸ. f current lags => power factor lags 3. f current leads => power factor leads 7

8 Power consumed nstantaneous power p=v*i = m Sinωt* m Sinωt = mm Sinωt Average power P pd t m m Sin t d t cos t P m m d t m m d t P cos t m m m m rms rms Waveform v i p Fig. c Note: t is also clear from the above graph that p is having some average value. Example 3: A 5 olts (MS), 5 Hz voltage is applied across a circuit consisting of a pure resistance of ohm. Determine i. The current flowing through the circuit ii. Power absorbed by circuit iii. Expression for voltage and current iv. Draw waveform and phasor diagram. Solution: 5 i. rms.5 Amp ii. Power absorbed ==5*.5=35 Watt iii. v 5 sin (5)t 5 sin t i.5 sin t iv. See figure c 8

9 . Purely nductive Circuit: Let applied voltage L v =m Sinωt () Fig. a ~ v When an alternating voltage is applied to purely inductive coil, an emf known as self induced emf is induced in the coil which opposes the applied voltage. Self induced emf di (t ) el L dt Since applied voltage at every instant is equal and opposite to self induced emf We have v=-el di (t ) m Sin t L dt di m Sin t dt L ntegrating Cos t m i m Sin t dt m Sin t / L L L Let m m L Here ωl=xl know as inductive reactance So i m Sin t / Here current lags by 9. Phasor diagram Φ =9, PF CosΦ=, Phase different between & is 9. Current () lags by 9 from voltage (). Power consumed nstantaneous power p=v*i = m Sinωt* m Sin(ωt-9) = -mm Sinωt Cosωt = -mm / Sinωt Average power Fig.b P pd t Ava m m Sin t Hence NO power is consumed in purely inductive circuit. 9

10 Waveform v i p Fig. c Note: t is also clear from the above graph that p is having ZEO average value. Example 4: The voltage and current thorough a circuit element v sin(34t 45 ) and i sin(34t 35 ) A. i. dentify the circuit elements ii. Find the value iii. Obtain expression for power Solution: i. Phase difference = =9 and lags v by 9, see the phasor diagram. So it is pure inductor Fig.3 ii. nductance XL=ωL => =34L => L=/34 H iii. p sin(34t 45 ) sin(34t 35 ) sin 68t 3. Purely Capacitive Circuit: Let applied voltage v =m Sinωt () C Fig. 4a ~ v are

11 Q Cv Q Cm Sin t dq d And we know i Cm Sin t dt m C Cos t dt dt i m Sin t / C Let m m C Here X C know as capacitive reactance C So i m Sin t / Here current leads by 9. Charge Phasor diagram Φ =9, PF CosΦ=, Phase different between & is 9. Current () leads by 9 from voltage (). Fig.4b Power consumed nstantaneous power p=v*i = m Sinωt* m Sin(ωt+9) = mm Sinωt Cosωt = mm / Sinωt Average power P pd t Ava m m Sin t Hence NO power is consumed in purely capacitive circuit. Waveform v i p Fig. 4c

12 Note: t is also clear from the above graph that p is having ZEO average value. Example 5: A 38 μf capacitor is connected to, 5 Hz supply. Determine i. Capacitive reactance offered by the capacitor ii. The maximum current iii. MS value of current drawn by capacitor iv. Expression for voltage & current Solution: i. XC=/ωC=/ᴫfC= Ohm ii. m=m/xc= /= A iii. rms = m/ = A iv. v sin t and i sin t / 4. -L Circuit: Let the applied voltage is oltage across resistance oltage across inductance L L v =m Sinωt () L + =. L=.XL =.(ωl) =. (πfl) - Fig. 5a - From phasor diagram L ( ) ( X L ) X L Z Z X L know as impedance Where From phasor diagram, is lagging behind by an angle ϕ X X X tan L L L tan L So i m Sin t Where m m Z Phasor diagram Phase different L ϕ. between & is Current () lags by ϕ from voltage (). & are in same phase, L is leading by an angle of 9 from etc. ϕ Fig.5b

13 Power consumed nstantaneous power p=v*i = m Sinωt* m Sin(ωt-ϕ) m m Sin tsin t m m cos cos t Average power P Ava m m cos cos t Ava m m cos Ava m m cos t m m Ava cos t P cos P m m cos cos Waveform v i p Fig. 5c oltage triangle and mpedance triangle: L Dividing each side by XL ϕ ϕ oltage Triangle Z Fig. 5d mpedance Triangle Power factor: Cos ϕ= /=/Z 3

14 Active power, reactive power, apparent power and power factor: Active power P = Cos 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 Example 6: A resistance and inductance are connected in series across a voltage v 83 sin(34t ) & current expression as i 4 sin(34t / 4). Find the value of resistance, inductance and power factor. Solution: Z=m/m=7.75 =Z cosϕ=7.75* cos(ᴫ/4)=5.7 ohm ωl => L =. ϕ=7.75* sin(ᴫ/4)=5.7 ohm = XL=Z sin 59 H Power factor = cos ϕ= cos(ᴫ/4)=/ =.77 Example 7: i 4.4 sin( t / 6) passes in an electric circuit when a voltage of v 4.4 sin t is applied to it. Determine the power factor of the circuit, the value of true power, apparent power and circuit components. Solution: Z=m/m=4.4/4.4= Ohm PF cosϕ=cos(ᴫ/6)=.866 P= cosϕ =(4.4/ )( 4.4/ )*.866=866 W S==(4.4/ )( 4.4/ )= W =Zcosϕ=*.866=8.66 XL=Zsinϕ=*.5=5 Ohm 5. -C Circuit: Let the applied voltage is v =m Sinωt () oltage across resistance oltage across inductance + - C C + C + =. Fig. 6a C=.XC =.(/ωc) =. (/πfc) - From phasor diagram 4

15 C ( ) ( X C ) X C Z Z X C know as impedance Where From phasor diagram, is leading by an angle ϕ X X X tan C C C tan C So i m Sin t Where m m Z Phasor diagram Phase different between & is ϕ. Current () leads by ϕ from voltage (). & are in same C phase, C is lagging by an angle of 9 from etc. ϕ Fig.6b Power consumed nstantaneous power p=v*i = m Sinωt* m Sin(ωt+ϕ) m m sin tsin t m m cos cos t Similar to -L circuit average power P cos Waveform v i p Fig. 6c 5

16 Note: Similar to -L circuit we can have voltage triangle and power triangle. 6. -L-C Circuit: C L + C C - + C - Fig. 7a Let the applied voltage is v =m Sinωt () oltage across resistance oltage across inductance oltage across inductance =. L=.XL C=.XC From phasor diagram L C ( ) ( X L X C ) X L X C Z Z X L X C know as impedance Where From phasor diagram, is lagging by an angle ϕ C X L X C X L X C X XC tan L tan L So i m Sin t Where m m Z Phasor diagram L L-C ϕ C 6 Fig.7b Assuming L>C

17 Power consumed Average power (similar to -L circuit) P cos Different cases:. L>C => nductive circuit, lags. L<C => Capacitive circuit, leads 3. L=C => esistive circuit, & are in same phase, power factor PF is Unity. This condition is called as electrical resonance. Example 8: A series LC circuit consisting of resistance of Ohm, inductance of. H and capacitance of 5 μf is connected across a 3, 5 Hz source. Calculate. Current. Magnitude & nature of power factor Solution: i. = Ohm XL=ᴫfL=ᴫ*5*.=6.8 Ohm XC=/ᴫfC=/(ᴫ*5*5*-6)=.3 Z X L X C ii. =/Z=4.98 A iii. PF=/Z=.433 lagging mpedance(z): Z X Where = esistance X= eactance PF: Cosϕ= /Z Ω, Ohm Ω Ω Z X ϕ mpedance Triangle Fig. 8 Admittance(Y): t is reciprocal of impedance Y=/Z Y G B Ω-, Mho, Siemens Where G= Conductance Ω- Y X= Susceptance Ω- PF: Cosϕ= G/Y ϕ G=YCos ϕ=(/z)(/z)=/z. ϕ=(/z)(x/z)=x/z B=YSin. G B AdmittanceTriangle 7 Fig. 9

18 Parallel Circuits: There are 3 methods for solving parallel circuits. Phasor Method: Consider the following circuit L C L C 3 L3 C Fig. a For branch Z X L X C Z Z if XL>XC (nductive circuit) (Leading PF) if XL<XC (Capacitive circuit) (Lagging PF) Similarly for branch and 3 we have Z,, Cosɸ and Z3, 3, Cosɸ3 respectively. cos Now we can draw the phasor diagram as shown bellow assuming any values for,, 3 and ɸ, ɸ, ɸ3 3 Φ3 Φ Φ Fig.b esultant current which will be the phasor sum of, & 3 can be determine by using any of following method i. Parallelogram method ii. esolving branch current, & 3 along x and y-axis. Component of resultant current along x-axis Cos Cos Cos 3 Cos 3 Component of resultant current along x-axis Sin Sin Sin 3 Sin 3 8

19 Sin Cos & tan Sin Cos. Admittance Method: Consider the following circuit Z, Y, G, B L C Z, Y, G, B L C Z, Y, G, B 3 L3 C Fig. Z X We know susceptance B Z Total conductance G=G+G+G3 Total susceptance B=B+B+B3 Total Admittance We know conductance G So => find out G, G & G3 => find out B, B & B3 Y G Y =Z => =/Z=Y =Y =Y 3=Y3 3. Symbolic or j-notation Method: Consider following voltage and current v m sin t i m sin t Wave form and phasor diagram will be as shown bellow 9

20 Fig. a Therefore either the time waveform of the rotating phasor or the phasor diagram can be used to describe the system. Since both the diagrams, the time diagram and the phasor diagram convey the same information, the phasor diagram being much more simpler, it is used for an explanation in circuit theory analysis. Since electrical data is given in terms of rms value, we draw phasor diagram with phasor values as rms rather than peak value used so far. Above voltage can be represented by = x + jy Cartesian Co-ordinates = θ in polar coordinates. Here x=component of along x-axis y= Component of along y-axis j is an operator which when multiplied 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 complex number. Note: n Phasor algebra: o Addition & Subtraction is done in Cartesian form. o Multiplication, Division, power and roots are done in polar form.

21 Phasor Algebra applied to single phase circuit. -L series circuit: L L L + - Fig. 3 j =+j L=+jXL =+L =( +j)+( +jxl) =+jxl =(+jxl) =Z Z=+jXL= Z =+j= Z Z Z Consider as reference oltage drop across resistance oltage drop across inductance So total voltage. Where Consider as reference -C series circuit: C C + C Fig. 4 Consider as reference oltage drop across resistance oltage drop across capatance So total voltage Where Consider as reference - j =+j L=-jXC =+C =( +j)+( -jxc) =-jxc =(-jxc) =Z Z=-jXC= Z =+j=

22 . Z Z Z Power determination: Let v=+3j volts & i=3-j Amp v i ( 3 j ) (3 j) 3 j Here Active power or real power = e v i =3 Watt eactive power = m v i = A -L-C parallel circuit: L C L C 3 L3 C Fig. 5 For branch : Z=+jXL j j Z jx L jx L jx L jx L X j L X L. X L Similarly for branch & 3 X j L X L X L X 3 3 j L 3 3 X L 3 3 X L 3 So total current =++3 Series-parallel circuit: mpedances in series Zeq=Z+Z+Z3+. mpedances in parallel

23 ... Z eq Z Z Z 3 A series parallel ac circuit can be solved in same manner as that of DC series parallel circuit except that complex impedances are used instead of resistances. Example 9: A 5mH inductor is in series with a parallel combination of a ohm resistance & μf capacitor. f ω of applied voltage is. Find i. Total impedance ii. Total admittance iii. Current in each branch if applied voltage is 3 Solution: =8Ω L=5mH C=μF Fig. 6 Z==8 Ω j 5 j C 6 Z3=jXL=jωL=jx5x-3 Ω =5jΩ Z=-jXC= j Equivalent impedance Z Z Z Z 3 i. ii. iii. 3 ZZ 8 5 j Z3 5 j Z Z 8 5 j Z = j = Y j Z j Z Z By current division rule Z Z 5 j j j Z By current division rule Z Z O =-= ( j )-( j )= j