DeCarlo Ch11 Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

This preview shows pages 1–6. Sign up to view the full content.

Complex Power Prbs 2/5/00 P11-1 @ DeCarlo & P. M. Lin CHAPTER 11. PROBLEM SOLUTIONS S OLUTION 11.1. Using equation 11.3, P av = 1 2 ( e t - 1) 2 Rdt = e 2 t 2 + t - 2 e t 0 1 = 0.758 0 1 W. S OLUTION 11.2. (a) From 11.6, P av = V m 2 2 R = 50 mW for a sinusoidal input. (b) From 11.3, P av = R 2 (10cos(10 t )) 2 dt 0 /20 + ( - 10cos(10 t )) 2 dt /20 3 /20 + (10cos(10 t )) 2 dt 3 /20 2 /10 = I m 2 R 2 = 50 mW just as the previous case since the square of the absolute cos(10t) is the same as the square of cos(10t). (c) P av = 10 R 2 0.01cos 2 (10 t ) [ ] 2 dt 0 2 /10 = 10 - 3 2 R 0.5cos(20 t ) + 0.125cos(40 t ) + 0.375 [ ] dt 0 2 /10 = 1 2 0.5 20 sin(20 t ) + 0.125 40 sin(40 t ) + 0.375 t 0 2 /10 = 37.5 mW (d) t=0:1/1000:1; R=1e3; pta= (0.01*cos(10.*t)).^2.*R; ptb= (0.01*abs(cos(10.*t))).^2.*R; ptc= (0.01*cos(10.*t)).^4.*R; subplot(3,1,1); plot(t,pta); grid ylabel('W'); subplot(3,1,2); plot(t,ptb); grid ylabel('W'); subplot(3,1,3); plot(t,ptc); grid ylabel('W'); xlabel('time in seconds');

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex Power Prbs 2/5/00 P11-2 @ DeCarlo & P. M. Lin 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 W TextEnd 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 W TextEnd 0 0.2 0.4 0.6 0.8 1 0 0.5 1 x 10 -5 W TextEnd time in seconds S OLUTION 11.3. (a) For figure a, the period is 2, and P av = 1 2 v 2 ( t ) R 0 2 dt = 1 2 R 400 + 100 ( 29 = 25 W. In figure b, the period is 1, and P av = 20 t ( 29 2 R 0 1 dt = 400 R t 3 3 0 1 = 13.3 W. (b)
Complex Power Prbs 2/5/00 P11-3 @ DeCarlo & P. M. Lin 0 0.5 1 1.5 2 2.5 3 3.5 4 10 15 20 25 30 35 40 Time in s Instantaneous Power in W TextEnd 0 0.5 1 1.5 2 2.5 3 3.5 4 0 5 10 15 20 25 30 35 40 Time in s Instantaneous Power in W TextEnd S OLUTION 11.4. (a) For (a), looking the definition for the effective voltage, one sees graphically that the integral over one period, 2, of the squared waveform, is 500. Dividing by the period, and taking the square root, V eff = 15.81 V. For fig. b, V eff = 20 t ( 29 2 dt 0 1 = 400 1 3 = 11.55 V (b) P av = I eff 2 R = V eff 10 2 8 = 20 W. (c) P av = V eff 10 2 8 = 10.67 W

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex Power Prbs 2/5/00 P11-4 @ DeCarlo & P. M. Lin S OLUTION 11.5. (a) This can be done graphically quite easily. The period of fig a, is 9s. The total area of one period of the squared waveform is 75. This yields I eff = 75 9 = 2.89 A. In fig. b, the area over one period is 25 which yields I eff = 25 3 = 2.89 A. (b) Using current division, P av = I eff 60 90 2 30 = 111.36 W. (c) The same result as (b) is obtained since the effective current is the same. S OLUTION 11.6. (a) I eff = 1 2 e t - 1 ( 29 2 0 1 dt = 0.615 A. (b) P av = I eff 2 R = 0.758 W. S OLUTION 11.7. (a) V eff 2 = 20 2 10 + 2cos(20 t ) ( 29 2 0 2 /20 dt = 20 2 100 t + 2 t + 2 40 sin(40 t ) + 2sin(20 t ) 0 2 /20 = 102.01 Hence V eff = 10.1 V. (b) V eff 2 = 1 10cos(2 t ) + 5cos(4 t ) ( 29 2 0 dt = 1 62.5 t [ ] 0 = 62.568 Hence V eff = 7.91 V. (c) Without going into detailed calculation, note the following fact about v 3 2 ( t ) . Only the product terms that have the same frequency will produce a non-zero result when integrated. Thus the integral reduces to the following: V eff 2 = 1 100cos 2 (2 t ) + 25cos 2 (4 t ) + 25cos 2 (4 t - π /4) + 50cos(4 t )cos(4 t - π /4) ( 29 0 dt Hence V eff = 50 + 12.5 + 12.5 + 25cos( / 4) [ ] = 9.627 V S OLUTION 11.8. The voltage is V = 50 0 V and the impedance Z eq = 100 - 100 j = 141.42 ∠ - 45 o . Thus I = V Z eq = 353.6 45 o mA. Hence, P av = R I eff 2 = 100 × 0.3536 2 2 = 6.2516 W.
Complex Power Prbs 2/5/00 P11-5 @ DeCarlo & P. M. Lin S OLUTION 11.9. (a)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern