Download presentation

Presentation is loading. Please wait.

1
Chapter 11 AC power analysis SJTU

2
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

3
**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

4
**AC power analysis Instantaneous Power Suppose: i(t) v(t) N**

Invariable part Sinusoidal part SJTU

5
E page415 figure 10.2 SJTU

6
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

7
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

8
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

9
**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

10
**Apparent power <0 or >0 S=VrmsIrms (VA) Power factor**

current leads voltage or current lags voltage < or >0 SJTU

11
**Reactive power (VAR) Resistor: Q=0 Inductor: Q=VrmsIrms**

Capacitor: Q=-VrmsIrms To any passive single port network SJTU

12
The power triangle S Q P SJTU

13
EXAMPLE Find the average power delivered to the load to the right of the interface in Figure 8-64. Fig. 8-64 SOLUTION: SJTU

14
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

15
**Are these equations right?**

SJTU

16
**Maximum power transfer**

Fig. 8-66: A source-load interface in the sinusoidal steady state. SJTU

17
**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

18
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

19
SJTU

20
**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

21
**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

22
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

Similar presentations

OK

SJTU1 Chapter 3 Methods of Analysis. SJTU2 So far, we have analyzed relatively simple circuits by applying Kirchhoffs laws in combination with Ohms law.

SJTU1 Chapter 3 Methods of Analysis. SJTU2 So far, we have analyzed relatively simple circuits by applying Kirchhoffs laws in combination with Ohms law.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on coastal plains of india Ppt on history of olympics video Ppt on features of ms excel Ppt on power system harmonics calculation Ppt on tata trucks india Ppt on iso-22000 and haccp Ppt on limits and continuity in calculus Ppt on ball bearing Ppt on business environment nature concept and significance level Ppt on moral education in schools