# Chapter 11 AC power analysis SJTU.

## Presentation on theme: "Chapter 11 AC power analysis SJTU."— Presentation transcript:

Chapter 11 AC power analysis SJTU

rms 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. effective value or DC-equivalent value The rms value of a periodic function is defined as the square root of the mean value of the squared function. SJTU

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. SJTU

AC power analysis Instantaneous Power Suppose: i(t) v(t) N
Invariable part Sinusoidal part SJTU

E page415 figure 10.2 SJTU

Stored energy WLav In the sinusoidal steady state an inductor operates with a current iL(t)=IAcos(wt). The corresponding energy stored in the element is Average stored energy SJTU

Stored energy In the sinusoidal steady state the voltage across a capacitor is vc(t)=VAcos(wt). The energy stored in the element is Average stored energy WCav SJTU

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) 1) 2) SJTU

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 Inductor case Pav = 0 Resistor case Average power Pav=0.5Vm*Im Pav=vrms*irms SJTU Capacitor case Pav = 0

Apparent power <0 or  >0 S=VrmsIrms (VA) Power factor
current leads voltage or current lags voltage < or  >0 SJTU

Reactive power (VAR) Resistor: Q=0 Inductor: Q=VrmsIrms
Capacitor: Q=-VrmsIrms To any passive single port network SJTU

The power triangle S Q P SJTU

EXAMPLE Find the average power delivered to the load to the right of the interface in Figure 8-64. Fig. 8-64 SOLUTION: SJTU

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 SJTU

Are these equations right?
SJTU

Maximum power transfer

we know P is maximized when RL=RT
Let XL=-XT then we know P is maximized when RL=RT the maximum average power where |VT| is the peak amplitude of the Thevenin equivalent voltage SJTU

EXAMPLE                                                                                   (a) Calculate the average power delivered to the load in the circuit shown in Figure 8-67 for Vs(t)=5cos106t, R=200 ohm, and RL=200 ohm. (b) Calculate the maximum average power available at the interface and specify the load required to draw the maximum power. SOLUTION: (a) SJTU

SJTU

If the load must be a resistor, how get the maximum power on it?
Question: If the load must be a resistor, how get the maximum power on it? SJTU

Maximum power transfer when ZL is restricted
RL and XL may be restricted to a limited range of values. In this situation, the optimum condition for RL and XL is to adjust XL as near to –XT as possible and then adjust RL as close to as possible the magnitude of ZL 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 ZL is set equal to the magnitude of ZT SJTU

Note: 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? If ZL cannot be varied but ZT can, what value of ZT results in maximum average-power transfer to ZL? SJTU