Presentation on theme: "SJTU1 Chapter 11 AC power analysis. SJTU2 rms value The rms value of a periodic function is defined as the square root of the mean value of the squared."— Presentation transcript:
SJTU1 Chapter 11 AC power analysis
SJTU2 rms value The rms value of a periodic function is defined as the square root of the mean value of the squared function. effective value or DC-equivalent value The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC (constant) value which gives the same effect.
SJTU3 If the periodic function is a sinusoid, then What do AC meters show, is it the RMS or peak voltage? AC voltmeters and ammeters show the RMS value of the voltage or current. What does '6V AC' really mean, is it the RMS or peak voltage? If the peak value is meant it should be clearly stated, otherwise assume it is the RMS value.
SJTU4 AC power analysis Instantaneous Power v(t) i(t) N Suppose: Invariable part Sinusoidal part
SJTU5 E page415 figure 10.2
SJTU6 Stored energy In the sinusoidal steady state an inductor operates with a current i L (t)=I A cos(wt). The corresponding energy stored in the element is Average stored energy W Lav
SJTU7 Stored energy In the sinusoidal steady state the voltage across a capacitor is v c (t)=V A cos(wt). The energy stored in the element is Average stored energy W Cav
SJTU8 Average power The average power is the average of the instantaneous power over one period real power Note :There are other methods to calculate P. 1) 2)
SJTU9 Instantaneous power, real power Instantaneous power waveforms for a voltage of 2V peak and a current of 1.5A peak Flowing separately in a resistor, a capacitor and an inductor Resistor case Average power P av =0.5V m *I m P av =v rms *i rms Inductor case P av = 0 Capacitor case Pav = 0
SJTU10 Apparent power S=V rms I rms (VA) Power factor current leads voltage or current lags voltage 0
SJTU11 Reactive power Resistor: Q=0 Inductor: Q=V rms I rms Capacitor: Q=-V rms I rms (VAR) To any passive single port network
SJTU12 The power triangle Q S P
SJTU13 EXAMPLE Find the average power delivered to the load to the right of the interface in Figure Fig SOLUTION:
SJTU14 Complex power Complex power is the complex sum of real power and reactive power =P+jQ So =VI* Where V is the voltage phsor across the system and I* is the complex conjugate of the current phasor. The magnitude of complex power is just apparent power
SJTU15 Are these equations right?
SJTU16 Maximum power transfer Fig. 8-66: A source-load interface in the sinusoidal steady state.
SJTU17 Let X L =-X T then we know P is maximized when R L =R T the maximum average power where |V T | is the peak amplitude of the Thevenin equivalent voltage
SJTU18 EXAMPLE (a) Calculate the average power delivered to the load in the circuit shown in Figure 8-67 for V s (t)=5cos10 6 t, R=200 ohm, and R L =200 ohm. (b) Calculate the maximum average power available at the interface and specify the load required to draw the maximum power. SOLUTION: (a)
SJTU20 (b) Question: If the load must be a resistor, how get the maximum power on it?
SJTU21 Maximum power transfer when Z L is restricted 1)R L and X L may be restricted to a limited range of values. In this situation, the optimum condition for R L and X L is to adjust X L as near to –X T as possible and then adjust R L as close to as possible 2)the magnitude of Z L can be varied but its phase angle cannot. Under this restriction, the greatest amount of power is transferred to the load when the magnitude of Z L is set equal to the magnitude of Z T
SJTU22 Note: 1.If the load is a resistor, then what value of R results in maximum average-power transfer to R? what is the maximum power then? 2.If Z L cannot be varied but Z T can, what value of Z T results in maximum average-power transfer to Z L ?