Microsoft PowerPoint - AC Ciruits

AC Circuits.

[Audio] Unidirectional-atanyspecifiedtimethevalueofthevoltageorcurrentisthesame MostcommonlyknownasDirectCurrentSignal.

[Audio]  Periodic- the voltage or current signals recur at a regular time interval  Aperiodic- the waveform has no recurrence.

[Audio]  Alternating Current Signal-the voltage or current is continuously changing as a function of time  electric charge periodically reverses direction.

Alternating Current Signals Sine Triangle Sawtooth.

[Audio] FlowofCurrentofDCandACDCcurrentflowisunidirectional. ACcurrentflowisbackandforth..

[Audio] Magnetic Lines of Flux. Magnetic Lines of Flux.

[Audio] AC Generator "Whenever lines of flux is cut by a conductor, emf is induced in it" -Michael Faraday(1791-1867).

[Audio] AC DC Cannot be stored Can be stored *portable gadgets Easier to produce High cost of materials to Cheaper than DC in produce DC transmitting Can be produced in bulk Voltage cannot be changed Voltage value can be change through the use of transformers.

[Audio]  The sine wave is the fundamental type of alternating current (ac) and alternating voltage. It is also referred to as a sinusoidal wave, or, simply, sinusoid. The electrical service provided by the power companies is in the form of sinusoidal voltage and current.  Sine waves are produced by two types of sources: rotating electrical machines (ac generators) or electronic oscillator circuits, which are used in instruments commonly known as electronic signal generators..

Sine Wave + voltage or + current - voltage or - current Positive maximum Time(t) Negative maximum.

[Audio] Asinewavevarieswithtime(t)inadefinablemanner. Thetimerequiredforagivensinewavetocompleteonefullcycleiscalledtheperiod (T)..

[Audio]  Frequency is the number of cycles that a sine wave completes in one second.  The more cycles completed in one second, the higher the frequency.  Frequency (f) is measured in units of hertz. One hertz (Hz) is equivalent to one cycle per second.

[Audio]  The relationship between frequency and period is important.  The formulas for this relationship are as follows:  There is a reciprocal relationship between f and T.

[Audio] The peak value of a sine wave is the value of voltage (or current) at the positive or the negative maximum (peaks) with respect to zero. Since the peaks are equal in magnitude, a sine wave is characterized by a single peak value, For a given sine wave, the peak value is constant and is represented by Vp or Ip..

[Audio] voltage (or current) from the positive peak to the negative peak. It is always twice the peak value as expressed in the following equations. Peak-to-peak values are represented by Vpp or Ipp. Vpp = 2Vp Ipp = 2Ip.

[Audio]  The average value of a sine wave is defined over a half-cycle rather than over a full cycle  The RMS value is also referred to as the T effective value. The RMS value is the equivalent heating effect of AC signal to that of DC m Y 2 Y Y m ave   p p Y  Y Y Y   eff m rms 2 1.

[Audio] Where Ym=max value of the V/I       t Y Y t m sin =2πf (rad/sec) =angular velocity f=frequency t=any time t(sec) = phase angle A  B  m Y m Y A B    A  m t Y Y t    sin A leading by A B lagging by B    B  m t Y Y t    sin.

[Audio] Given:    o  t v t 45 150sin 1000   V V 150  m V m 150 V V .07 106    eff 2 2 V V V 300 2   m p p  Find: Vm Veff Vpp Vave 2  Frequency Period Draw the waveform V V V .49 95   m ave  1000 rps   45o  V 150 Hz f 159    2 2 1000    3 1 x period 28 10 sec .6   f V 300 28ms .6.

[Audio] 1. Resistor V I V=iR the voltage and the current waveforms are in phase    t V v t m sin      t i t R Vm sin  .

[Audio] 2. Inductor- stores energy in each magnetic field -inductance (L) unit is Henry(H)  N S .

[Audio] Inductor V L L V  dt di I V i  L L L dt t V cos   m  L  V m 90 sin t     o L   Voltage leads the current by 90o Current lags the voltage by 90o.

[Audio] 3. Capacitor - stores energy in each electric field - capacitance(C) unit is Farad (F) plates            Gap or insulation    .

[Audio] Capacitor C i V   C dt V C i  C dt dv I t V d sin    m C  dt t CV 90 sin       o m Current leads the voltage by 90o Voltage lags the current by 90o.

[Audio] Capacitor Inductor 1 1 1 1 n T L L L L    2 1 Series n T C C C C 2 1    1 1 1 1 Parallel n T C C C C    2 1 n T L L L L 2 1   .

[Audio] Capacitor Inductor Resistor V m C t CV i t 90 sin     L t t i L m 90 sin         o    t i t R Vm sin       o C t t v C I     o m 90 sin        t RI v t m sin   m L t LI v t 90 sin         o 1 1 j j jX j L fL j jX 2     C fC C 2       L R o o X 90   X 90   L C Where R=resistance Xc =capacitive reactance XL =inductive rectance.

[Audio] j  jX R Z o    Z X 2 2 2 X R Z    R R tan 1 X          R j  Where Z= impedance R= resistance X=reactance Θ=impedance angle If the reactance (X) is positive, Z is inductive in nature. If the reactance (X) is positive, Z is inductive in nature. Else, Z is capacitive in nature.

[Audio] R-L Circuit R-C Circuit R Z L X XC   R R Z R o o jX R Z     jXC  L jX R Z      C L jX L C  tan1  R X  tan1  R X  Xc is negative  XL is positive  Theta (θ) is negative  Theta (θ) is positive.

[Audio] L-C Circuits If the total reactance is jXC  negative, the load is more capacitive. C L T X X X   If the total reactance is positive the load is more inductive L jX Capacitive Inductive o o T jX X  90 T T T jX X     90.

[Audio] j  Where i j 1    2 Z X j 1    3 R j j   R 4 j 1  j              Z Zcis j Z jX R sin cos.

[Audio] 1 1 1 jX R Z   2 2 2 jX R Z   1. Addition/ Subtraction Z Z Z 2 1 3 X j X R R       *for polar: change to rectangular form then add or subtract    2  1 2 1 2. Multiplication jX R jX R Z Z Z        2 2 1 1 2 1 3 X R j X R X X R R        1  2 2 1 2 1 2 1  2  1 2 1 3 3        Z Z Z.

[Audio] 3. Division jX R     1 1 1 2 2 jX R Z Z Z 3 2 jX R jX R            2 2 2 2 X R j X R X X R R     1 2 2 1 2 1 2 1     X R  2 2 2 2 1  2  1 3 3        Z Z Z 2.

[Audio] j Z 18 12   1 j Z 15 4   2 Z 25  3 15 18 4 12 Z Z    2 1 4 j j Z Z Z      2 1 15 4 18 12 j j    180 72 270 48 j j 108 318     3 16 j 3 16 j j     108 318 3 16 2682 4764 j     3 16 j j 3 16 j j 265          10.12 .98 17 j   .

[Audio] 10.12 (25) .98 17 Z Z   3 4 ab 3 4 j j Z Z Z 25 10.12 .98 17       253 5. 449 253 5. 449 10.12 .98 42   10.12 .98 42 j j 10.12 .98 42 j j 10.12 .98 42 j j               6330.61 .38 21858 j   .69 1949 .3 25 .21 11 j   .

[Audio] 10.33 a 100Ω resistor and a 500µF capacitor are in series. The voltage across the resistor is Vr=300sin 100t. Determine the (a) current, (b) the voltage across the capacitor and (c)impedance sin100 300 t (a)   R V I R 100 sin100 3 tA  Phasor form o o 300 V V     R 0 212 1. 0 2 o 0 1. 212 o A I      R V R .2 12 0 100.

[Audio] 1       (b) (500 ) 100 j j jX C C  j 20    j jX I V C C C I      o 90 3sin 100 20 t       o 90 sin 100 60    V t Phasor form o o X I V .2 12 0 90 20         C C o 90 4. 42 V      (c) jX R Z C t 20 100 j   .

[Audio] 11.60 Determine the voltages across each impedance o   V   5 60 120 1 o o o 40 15 10 30 60 5           o   5 60 120   o 120 .0 22 60.79 .0 79 .65 22           60.79 4. 26 V  .

[Audio] o   10 30 120 2 V o .0 79 .65 22         .0 44 30.79 120     30.79 8. 52 V   o     40 15 120 3 V o .0 79 .65 22       39.21 .0 66 120      39.21 2. 79 V   .

Example 3 120LOOV IOL300Q.

[Audio] P=RealPower,WattsZXfunctionofresistance,R  Q=ReactivePower,VAR(VoltRAmpere-Reactance)functionofreactance,XS=ApparentPower,VA(VoltSQ  Ampere)functionofimpedance,ZP.

[Audio] jX R Z o    2 2 2 X R Z   Z X  Angle of Impedance         R tan 1 X  R jQ P S o    S Q 2 2 2 Q P S     P P Q         tan 1 Power Factor Angle *phase angle difference between the voltage and current S P cos   S Q sin   cos   Power Factor S P.

[Audio] 2 �𝑅 I Z S  𝑃 = 𝐼� Z � 2 = 𝑉� Z 𝑅 Watts VA  = 𝑉 𝐼 cos 𝜃 = 𝑉� 𝐼� V I Z V  V *I S pf  2 I X Q  X 2 X  VAR, inductive (lagging pf) VAR, capacitive (leading pf) V I X V   sin S Q  P.

[Audio] Where: θ= phase angle difference of the voltage and current (θV –θI ) cos    pf = Power Factor or the ratio of the S P pf real power to the apparent power sin    rf =Reactive Factor or the ratio of the S Q rf reactive power to the apparent power Power Factor/ Reactive Factor: Lagging- inductive Leading- capacitive Unity- resistive (cosθ=1 or sinθ=0).

Example 3 120LOOV IOL300Q.

[Audio] 11.89Theloadatacertainfactoryconsistsofthefollowing:(i)10kW,pf=0.6lagging (ii)15kW,rf=0.5lagging,(iii)25kVA,pf=0.75lagging(iv)22kVA,rf=0.65lagging(v)15kvarpf=0.8laggingand(vi)5kVAR,purecapacitance.(a)Determinethetotaltruepower,totalreactivepowerandthetotalapparentpower.

[Audio] o 53.13 6.0 cos 1     (i) pf=0.6 lagging   P=10kW Q  13.33 10 53.13 tan kVAR Q  o 30 5.0 sin 1     (ii) rf=0.5 lagging   P=15kW Q  .8 66 15 30 tan kVAR Q .

[Audio] o 41.41 .0 75 cos 1       (iii) pf=0.75 lagging P S=25kVA Q   16.54 25 41.41 sin kVAR Q  18.75 25 41.41 cos kW P  o 40.54 .0 65 sin 1     (iv) rf=0.65 lagging   S=22kVA P Q   14 3. 22 40.54 sin kVAR Q  16.72 22 40.54 cos kW P .

[Audio] o (v) pf=0.8 lagging 36.87 8.0 cos 1       Q=15kVAR 15 36.87 tan  P kW P 20  (vi) Q= 5kVAR pure capacitance.

[Audio] P P P P P P P       T 6 5 4 3 2 1 0 20 15 18.75 16.72 10       .47 80 kW Q Q Q Q Q Q Q        T 6 5 4 3 2 1 5 15 .8 66 16.54 14 3. .33 13       .83 62 kVAR  2 2 2 2 Q P S 62.83 .47 80     T T T .09 102 kVA .

[Audio] The power factor is to be corrected to 0.9 lagging. (b) determine the kVAR rating of pure capacitors needed to accomplish this.   37.98o tan .47 80 1 62.87       25.84o 9.0 cos 1     .47kW 80 25.84o .98o 37 .83kVAR 62 .09kVA 102 C Q.