4 Average PowerAverage Power, in watts (W), is the average of instantaneous power over one period
5 Average Power Resistive load (R) absorbs power all the time. For a purely resistive circuit, the voltage and the current are in phase (v = i).
6 Average Power Reactive load (L or C) absorbs zero average power. For a purely reactive circuit, the voltage and the current are out of phase by 90o (v - i = ±90).
7 Exercise 11.3 Find the average power supplied by the source and the average power absorb by the resistor
8 Solution The current I is given by The average power supplied by the voltage source is
9 Solution The current through the resistor is The voltage across resistor isThe average power absorbed by the resistor isNotice that the average power supplied by the voltage source is same as the power absorbed by the resistor.This result shows the capacitor absorbed zero average power.
10 Practice Problem 11.3 Calculate the average power absorbed by the resistor and the inductor. Then find the average power supplied by the voltage source
11 SolutionThe current I is given byFor the resistor
12 Solution For the inductor The average power supplied by the voltage source isNotice that the average absorbed by the resistor is same as the power supplied by the voltage source.This result shows the inductor also absorbed zero average power.
14 Maximum Power Transfer For maximum power transfer, the load impedance ZL must equal to the complex conjugate of the Thevenin impedance Zth
15 Maximum Average Power The current through the load is The Maximum Average Power delivered to the load is
16 Maximum Average PowerBy setting RL = Rth and XL = -Xth , the maximum average power isIn a situation in which the load is purely real, the load resistance must equal to the magnitude of the Thevenin impedance.
17 Exercise 11.5 Determine the load impedance ZL that maximize the power drawn and the maximum average power.
18 Solution First we obtain the Thevenin equivalent To find Zth, consider circuit (a)To find Vth, consider circuit (b)
19 SolutionFrom the result obtained, the load impedance draws the maximum power from the circuit whenThe maximum average power is
20 Practice Problem 11.5 Determine the load impedance ZL that absorbs the maximum average power. Calculate the maximum average power.
21 Solution First we obtain the Thevenin equivalent To find Zth, consider circuit (a)To find Vth, consider circuit (b)By using current divider
22 SolutionFrom the result obtained, the load impedance draws the maximum power from the circuit whenThe maximum average power is
23 Example 11.6 Find the value of RL that will absorbs maximum average power. Then calculate that power.
24 Solution First we obtain the Thevenin equivalent Find Zth Find Vth By using voltage divider
25 Solution The value of RL that will absorb the maximum average power is The current through the load isThe maximum average power is
26 Practice Problem 11.6 Find the value of RL that will absorbs maximize average power, Then calculate the power.
27 Solution First we obtain the Thevenin equivalent To find Zth let and ThenTo find VthBy using voltage divider
28 Solution The value of RL that will absorb the maximum average power is The current through the load isThe maximum average power is
30 Complex Power Apparent Power, S (VA) Real Power, P (Watts) Reactive Power, Q (VAR)Power Factor, cos
31 Complex PowerComplex power is the product of the rms voltage phasor and the complex conjugate of the rms current phasor.Measured in volt-amperes or VAAs a complex quantityIts real part is real power, PIts imaginary part is reactive power, Q
34 Complex Power (Derivation) From derivation, we notice that the real power isorand also the reactive poweror
35 Real or Average PowerThe real power is the average power delivered to a load.Measured in watts (W)The only useful powerThe actual power dissipated by the load
36 Reactive PowerThe reactive power, Q is the imaginary parts of complex power.The unit of Q is volt-ampere reactive (VAR).It represents a lossless interchange between the load and the sourceQ = 0 for resistive load (unity pf)Q < 0 for capacitive load (leading pf)Q > 0 for inductive load (lagging pf)
37 Apparent PowerThe apparent power is the product of rms values of voltage and currentMeasured in volt-amperes or VAMagnitude of the complex power
38 Power FactorPower factor is the cosine of the phase difference between voltage and current.It is also cosine of the angle of the load impedance.
39 Power Factor The range of pf is between zero and unity. For a purely resistive load, the voltage and current are in phase so that v- i = 0 and pf = 1, the apparent power is equal to average power.For a purely reactive load, v- i = 90 and pf = 0, the average power is zero.
40 Power TriangularComparison between the power triangular (a) and the impedance triangular (b).
41 Problem 11.46For the following voltage and current phasors, calculate the complex power, apparent power, real power and reactive power. Specify whether the pf is leading or lagging.V = 22030o Vrms, I = 0.560o Arms.V = 250-10o Vrms, I = 6.2-25o Arms.V = 1200o Vrms, I = 2.4-15o Arms.V = 16045o Vrms, I = 8.590o Arms.
42 Solution a) S = VI* = (22030o)( 0.5-60o) = 110-30o VA = – j55 VAApparent power = 110 VAReal Power = WReactive Power = -55 VARpf is leading because current leads voltageb) S = VI* = (250-10o)(6.225o)= 155015o VA = j401.2 VAApparent power = 1550 VAReal Power = WReactive Power = VARpf is lagging because current lags voltagec) S = VI* = (1200o)( 2.415o)= 28815o VA = j74.54 VAApparent power = 288 VAReal Power = WReactive Power = VARpf is lagging because current lags voltaged) S = VI* = (16045o)(8.5-90o)= 1360-45o VA = – j961.7 VAApparent power = 1360 VAReal Power = WReactive Power = VARpf is leading because current leads voltage
43 Problem 11.48 Determine the complex power for the following cases: P = 269 W, Q = 150 VAR (capacitive)Q = 2000 VAR, pf = 0.9 (leading)S = 600 VA, Q = 450 VAR (inductive)Vrms = 220 V, P = 1 kW, |Z| = 40 (inductive)
44 Solution Given P = 269W, Q = 150VAR (capacitive) Complex power, b) Given Q = 2000VAR, pf = 0.9 (leading)Complex power,
45 Solutionc) Given S = 600VA, Q = 450VAR (inductive)Complex power,
46 Solution d) Given Vrms = 220V, P = 1kW, |Z| = 40 (inductive) Complex power,
47 Problem 11.42A 110Vrms, 60Hz source is applied to a load impedance Z. The apparent power entering the load is 120VA at a power factor of lagging. CalculateThe complex powerThe rms current supplied to the load.Determine ZAssuming that Z = R + j L, find the value of R and L.
48 Solution Given S = 120VA, pf = 0.707 = cos = 45o a) the complex powerb) the rms current supplied to the load
49 Solution c) the impedance Z d) value of R and L If Z = R + jL then Z = j
50 Problem 11.83Oscilloscope measurement indicate that the voltage across a load and the current through is are 21060o V and 825o A respectively. DetermineThe real powerThe apparent powerThe reactive powerThe power factor
51 Solution a) the real power b) the apparent power c) the reactive power d) the power factor
53 Power Factor Correction The process of increasing the power factor without altering the voltage or current to the original load.It may be viewed as the addition of a reactive element (usually capacitor) in parallel with the load in order to make the power factor closer to unity.
54 Power Factor Correction Normally, most loads are inductive. Thus power factor is improved or corrected by installing a capacitor in parallel with the load.In circuit analysis, an inductive load is modeled as a series combination of an inductor and a resistor.
56 Calculation If the original inductive load has apparent power S1, then P = S1 cos 1 and Q1 = S1 sin 1 = P tan 1If we desired to increased the power factor from cos1 to cos2 without altering the real power, then the new reactive power isQ2 = P tan 2The reduction in the reactive power is caused by the shunt capacitor is given byQC = Q1 – Q2 = P (tan 1 - tan 2)
57 CalculationThe value of the required shunt capacitance is determined by the formulaNotice that the real power, P dissipated by the load is not affected by the power factor correction because the average power due to the capacitor is zero
58 Example 11.15When connected to a 120V (rms), 60Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.
59 Solution If the pf = 0.8 then, cos1 = 0.8 1 = 36.87o where 1 is the phase difference between the voltage and current.We obtained the apparent power from the real power and the pf as shown below.The reactive power is
60 Solution When the pf raised to 0.95, cos2 = 0.95 2 = 18.19o The real power P has not changed. But the apparent power has changed. The new value isThe new reactive power is
61 SolutionThe difference between the new and the old reactive power is due to the parallel addition of the capacitor to the load.The reactive power due to the capacitor isThe value of capacitance added is
62 Practice Problem 11.15Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume the load is supplied by a 110V (rms) 60Hz power line.
63 Solution If the pf = 0.85 then, cos1 = 0.85 1 = 31.79o where 1 is the phase difference between the voltage and current.We obtained the apparent power from the reactive power and the pf as shown below.The real power is
64 Solution When the pf raised to 1 (unity), cos2 = 1 2 = 0o The real power P has not changed. But the apparent power has changed. The new value isThe new reactive power is
65 SolutionThe difference between the new and the old reactive power due to the parallel addition of the capacitor to the load.The reactive power due to the capacitor isThe value of capacitance is
66 Problem 11.82A 240Vrms, 60Hz source supplies a parallel combination of a 5 kW heater and a 30 kVA induction motor whose power factor is DetermineThe system apparent powerThe system reactive powerThe kVA rating of a capacitor required to adjust the system power factor to 0.9 laggingThe value of capacitance required
67 Solution For the heater P1 = 5000 Q1 = 0 For the 30kVA induction motor, the pf = 0.82 then,cos1 = 1 = 34.92oThe real and the reactive power for the induction motor
68 Solution The total system complex power Stotal = S1 + S2 = (P1 + P2) + j (Q1 + Q2) = j17171The system apparent powerS = |Stotal| = 34.33kVAThe system reactive powerQ = kVARThe system power factor
69 Solution The system pf = 0.865 then, cos1 = 0.865 1 = 30.12o The new system pf = 0.9 then,cos2 = 0.9 2 = 25.84oThe rating for the capacitance required to adjust the power factor to 0.9QC = P (tan 1 + tan 2) = (tan tan 25.84) = 2833kVAR
71 Problem 11.91The nameplate of an electric motor has the following information.Line voltage: 220 VrmsLine current: 15 ArmsLine frequency: 60 HzPower: 2700 WDetermine the power factor (lagging) of the motor.Find the value of the capacitance, C that must be connected across the motor to raise the pf to unity.
72 Additional ProblemThe nameplate of an electric motor has the following information.Line voltage: 230 VrmsLine current: 13 ArmsLine frequency: 60 HzReactive Power: 1500 VARDetermine the power factor (lagging) of the motor.Find the reactive power required to increase the pf to 0.95.Find the value of the capacitance, C.