And the rms current is Irms = VArms/Vrms = 2.3 A. Problem 1-41 Part(a): Area increases by a factor of 4. Thus the voltage increases by a factor of 4 to e = 1096 cos(377t). C©2014 by McGraw-Hill Education. Ps3 cech 3008b proshivka pc. This is proprietary material solely for authorized instructor use.

May 14, 2013  Buenas noches con todos alguien tiene el solucionario de este libro de fitzgerald? La verdad que he buscado y buscado y no encuentro, seria de gran ayuda. Me gusta 0 0 No me gusta Responder 7:04 pm. Solucionario Mecanica De Materiales Fitzgerald Edi 11. Ozawa Deguchi Formato: PDF Espaol 496 Pags 18 MB Descargar libro.

Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Part (b): lc doubles therefore so does the current. Thus I = 0.26 A. Part (c): Volume increases by a factor of 8 and voltage increases by a factor of 4. There Iφ,rms doubles to 0.20 A. Part (d): Volume increases by a factor of 8 as does the core loss.

Thus Pc = 128 W. Problem 1-42 From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately) B = 0.47 T and H = -360 kA/m. Thus the maximum energy product is 1.69 ×105 J/m3. Because the neodymium-iron-boron magnet is essentially linear over the operating range of this problem, the system is linear and hence a sinusoidal flux variation will correspond to a sinusoidal current variation. Part (a): From the solution to Problem 1-46, the maximum energy product for neodymium- iron-boron at 180 C occurs at (approximately) Bm = 0.47 T and Hm = -360 kA/m.

The magnetization curve for neodymium-iron-boron can be represented as Bm = µRHm +Br where Br = 0.94 T and µR = 1.04µ0. The magnetic circuit must satisfy Hmd+Hgg = 0; BmAm = BgAg For Bg = 0.8 T, the minimum magnet volume will occur when the magnet is operating at the maximum energy point. Under these conditions, the source voltage will see a total impedance of Ztot = 1.5+j1.5 kΩ whose magnitude is 1.5√ 2 kΩ. The current will thus equal I = Vs/ Ztot = 5.7√ 2 mA.

Thus, the power delivered to the load will equal Pload = I2(N2Rload) = 48 mW Here is the desired MATLAB plot: c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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