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PowerPoint Presentation Fundamentals of Electric Circuits AC Circuits Chapter 11 AC power analysis 11 1 Introduction 11 2 Instantaneous and average power 11 3 Maximum average power transfer 11 4 Effec[.]

Fundamentals of Electric Circuits AC Circuits Chapter 11 AC power analysis 11.1 Introduction 11.2 Instantaneous and average power 11.3 Maximum average power transfer 11.4 Effective or RMS values 11.5 Apparent power, power factor and complex power 11.6 Conservative of AC power 11.7 Power factor correction 11.8 Applications FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.1 Introduction + In this chapter:  concern in power analysis + Power:  the most important quantity in electric utilities, electronic and communication system o Such systems involves transmission of power from one point to another o Each industrial and household electrical device has a power rating (how much power the equipment requires) FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.2 Instantaneous and average power + The instantaneous power p(t) absorted by an element:  product of the voltage v(t) across the element and the current i(t) through it p(t)  v(t).i(t) + Under sinusoidal excitation: v(t)  Vm cos(t  v ), i(t)  I m cos(t i ) p(t)  v(t)i(t)  1 VmI m cos(v  i )  VmI m cos(2t  v  i ) 2 part Part 1: i(t) Sinusoidal source part Always constant (time independent) Depend on the phase difference between v(t) and i(t) Part 2: Sinusoidal function Positive: power is absorbed by the source Negative: power is transferred from the circuit to the source (L, C) + v(t) - Passive linear network FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.2 Instantaneous and average power + The instantaneous power changes with time  difficult to measure  using the average power + Average power:  average of the instantaneous power over one period P We have: 1T V I cos(   ) p(t)dt  m m v i  T   VIˆ  Vm e j v I m e  ji  Vm I m e j  v i   Vm I m cos v   i   j sin  v   i   P  Re VIˆ For purely resistive circuit: For purely reactive circuit: 1 2 P  Vm I m  R I  RI 2 P  Vm I m cos 900    A resistive load (R) absorbs power at all times, while a reactive load (L, C) absorb zero average power FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.2 Instantaneous and average power + Example 1: Find the power generated by each source and the average power absorbed by each passive element R 20Ω 40 A Applying mesh current method:  I1  A   10.5879.10 A  I    Z I  60300   Z  Z I C L  L + + V1 - V2 C -j5Ω - I1 L I j10Ω 6030 0V Calculate average power of sources: PVsource      1 Re VIˆ2  Re 60300 10.58  79.10  207.8W (The circuit is delivering power to the voltage source) 2 V1  RI1  Z L I1  I2   184.986.210 V 1 PIsource   Re V1 Iˆ1   Re 184.986.210  400  367.8W (The current source is supplying power to the circuit) 2     FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.2 Instantaneous and average power + Example 1: Find the power generated by each source and the average power absorbed by each passive element R 20Ω 40 A Calculate average power absorbed by passive elements: 1 PR  RI12  20   160W 2 + + V1 - Sum of average power: PVsource  PIsource  PR  PL  PC  207.8  367.8  160    0W The total power supplied by the current source equals the power absorbed by the resistor and the voltage source V2 C -j5Ω - I1 L I j10Ω 6030 0V FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.3 Maximum average power transfer + In chapter 4:  the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance + Extend that result to AC circuit: ZTh  RTh  jXTh ZL  RL  jXL V Th V Th I   ZTh  ZL (RTh  jXTh )  (RL  jXL ) ZTh The average power delivered to the load: RL VTh 1 P  RL I  2 RTh  RL 2   X Th  X L 2 Z L  RL  jX L  RTh  jX Th  ZˆTh  P  R   max imum L P  0  X L VTh I ZL FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.3 Maximum average power transfer + Maximum average power transfer theorem for sinusoidal steady state: For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance ZTh VTh PMax  8RTh + If ZL = RL: the condition for maximum power transfer becomes 2 RL  RTh  XTh  ZTh FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.3 Maximum average power transfer + Example 2: For the given circuit as the next figure, find ZL that maximizes the average power What is the maximum average power? Find the Thevenin equivalent network at the load terminal R R  Z C  48  j  Z th  Z L   j5   2.93  j 4.47 R1  R2  Z C   j6 E R2  Z C   Vth   7.45  10.30 V R1  R2  Z C To have the maximum average power transfer, the load impedance is: Z L  Zˆ th  2.93  j 4.47 The maximum average power transfer to the load is: Pmax Vth 7.452    2.37W Rth  2.93 R1 L j5Ω 4Ω R2 8Ω 1000 V -j6Ω C ZL FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.3 Maximum average power transfer + Example 3: Find RL that will absorb the maximum average power Calculate that power? Find the Thevenin equivalent network at the load terminal Z L R  Z C   9.412  j 22.35 Z L  R  ZC E Z L  Vth   72.761340 V R  ZC  Z L Z th  RL  Z th  9.412  22.352  24.25  V 72 76  134 th   1.8100.20 A The current through the load is I  Z th  RL 33.39  j 22.35 Pmax  -j30Ω R C 15030 0V The value of RL that will absorb the maximum average power is The maximum average power absorbed by RL : 40Ω 2 I RL  39.29W j20Ω L RL FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.4 Effective or RMS values + Effective value: need to measure the effectiveness of a voltage or current source in delivering power to a resistive load + Effective value (rms - root mean square) of a periodic current is the DC current that delivers the same average power to a resistor as the periodic current I T  i (t)dt ; V  T0 T v  (t)dt for sinusoidal   I signal T0 Im ; V Vm 2 V P  RI  R + Voltmeter, ampe-meter, power-meter: are designed to read the rms value of voltage, current, and power, respectively FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.5 Apparent power, power factor and complex power + Apparent power [VA]: product of the rms values of voltage and current S  V I P P  V I cos v   i   S cos v   i   cos v   i    pf S V V v V Z    v   i I I i I (power factor) + Power factor: cosine of the phase difference between voltage and current  It is also the cosine of the angle of the load impedance power factor is said to be leading (capacitive load) or lagging (inductive load) FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.5 Apparent power, power factor and complex power + Complex power [VA]: product of the rms voltage phasor and the complex conjugate of the rms current phasor  its real part is real power P and its imaginary part is reactive power V ~  ˆ S  Vrms I rms  P  jQ  ZI rms  rms Zˆ + For a given load: jQ  Complex power contains all relevant power information  Power triangle, impedance triangle (voltage or current triangle) are always similar φ jXL L P φ ~ S R Z -jQC -jXC Power impedance Impedance triangle FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.6 Conservative of AC power + The principle of conservation of power applies to AC circuits as well as to DC circuits: The total power supplied by the source equals to the total power delivered to the load n  k  P P  load sources  n ~  k1 k Ssources   Sload   n k1 k Q sources   Qload  k1 The complex, real, and reactive powers of the sources equal the respective sums of the complex, real and reactive powers of the individual loads FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.6 Conservative of AC power + Example 4: Find the real and reactive power absorbed by the source, the line and the load The total impedance: Z  Rline  Rload  Z Lline  Z Cload  19  j8 j2Ω Rline Lline 15Ω Rload -j10Ω 22000 V Source The power delivered by the source: 4Ω Cload Line Load  I  VS  10.6722.830 A  S~  Iˆ  2347.4  22.830  2163.5  j 910.8VA  V source rms rms Z The power absorbed by the line: ~ Vline  Rline  Z Lline I  47.7249.40 V  Sline  Vline Iˆ  509.226.57  455.5  j 227.7VA The power absorbed by the load: ~ Vload  Rload  Z Cload I  192.38  10.87 V  Sload  Vload Iˆ  2053  33.7  1708  j1138.5VA FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.7 Power factor correction + Most domestic loads (washing machine, air conditioner, refrigerators, …) and industrial loads (induction motors) are inductive and operate at a low lagging power factor IC IC C IL I V I Inductive load + - + V 2 I V 1 Inductive load - + The process of increasing the power factor without altering the voltage or current to the original load  power factor correction + The real power P dissipated by the load is not affected by the power factor correction because the average power due to the C is zero FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 11.8 Applications Wattmeter: consists of coils + Current coil: Very low inpedance Connect in series with the load The + terminal is toward the source + Voltage coil: Very high impedance Connect in parallel with the load The + terminal is connected to the same line as the current coil Note: Reactive power is measured by an instrumentation called the VAmeter