1. Lecture 51 EEE 302 Electrical Networks II Dr. Keith E. Holbert Summer 2001

2. Lecture 52 Instantaneous Power: p(t) For AC circuits, the voltage and current are v(t) = V M cos(  t+  v ) i(t) = I M cos(  t+  i ) The instantaneous power is simply their product p(t) = v(t) i(t) = V M I M cos(  t+  v ) cos(  t+  i ) = ½V M I M [cos(  v -  i ) + cos(2  t+  v +  i )] Constant Term Wave of Twice Original Frequency

3. Lecture 53 MATLAB Exercise Start MATLAB at your computer Create and plot 60 Hz current & voltage waves EDU» t=0:0.0005:0.04; EDU» v=12*cos(377*t+pi/4); EDU» i=5*cos(377*t+pi/2); EDU» plot(t,v,'b:',t,i,'g--') Create and plot the instantaneous power from I and V EDU» p=v.*i; EDU» plot(t,v,'b:',t,i,'g--',t,p,'r') Don’t exit MATLAB, we’ll come back to it later

4. Lecture 54 Average Power (P) Calculate average power (integrate power over one cycle and divide by period) Recall that passive sign convention says: P > 0, power is being absorbed P < 0, power is being supplied

5. Lecture 55 Average Power: Special Cases Purely resistive circuit: P = ½ V M I M The power dissipated in a resistor is Purely reactive circuit: P = 0 –Capacitors and inductors are lossless elements and absorb no average power –A purely reactive network operates in a mode in which it stores energy over one part of the period and releases it over another part

6. Lecture 56 Class Examples Extension Exercise E9.1 Extension Exercise E9.2 MATLAB: compute the average power of the waveforms from earlier in the class

7. Lecture 57 Summarizing Does the expression for the resistor power look identical to that for DC circuits?

8. Lecture 58 Class Examples MATLAB EDU» pavg=12*5*cos(pi/4-pi/2)*0.5 Extension Exercise E9.3 Extension Exercise E9.4