Fundamentos de Circuitos Electricos - Sadiku - 3ed solucionario.pdf | Visual Cortex | Series And Parallel Circuits

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http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS. Chapter 1, S
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  http://www. elsolucionario.blogspot  .com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROSLOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRORESUELTOS Y EXPLICADOSDE FORMA CLARAVISITANOS PARADESARGALOS GRATIS .  Chapter 1, Solution 1 (a)   q = 6.482x10 17  x [-1.602x10 -19  C] = -0.10384 C (b) q = 1 . 24x10 18  x [-1.602x10 -19  C] = -0.19865 C (c) q = 2.46x10 19  x [-1.602x10 -19  C] = -3.941 C (d) q = 1.628x10 20  x [-1.602x10 -19  C] = -26.08 C Chapter 1, Solution 2 (a)   i = dq/dt = 3 mA (b)   i = dq/dt = (16t + 4) A (c)   i = dq/dt = (-3e -t + 10e -2t ) nA (d)   i=dq/dt = 1200 120 π π  cos  t   pA (e)   i =dq/dt = − + − e t  t 4 80 50 1000 50 ( cos sin )  A µ  t    Chapter 1, Solution 3 (a)   C 1)(3t  +=+= ∫ q(0)i(t)dtq(t) (b)   mC 5t)(t 2 +=++= ∫ q(v)dts)(2tq(t) (c)   ( ) q(t)20 cos 10t/6q(0)(2sin(10/6)1)C t  π π µ  = + + = + + ∫  (d)   C40t)sin0.12t(0.16cos40e 30t- +−=−+=+= ∫  t)cos40-t40sin30( 1600900e10q(0)t40sin10eq(t) -30t30t-   Chapter 1, Solution 4 ( ) mC4.698 =−=−=== ∫ ∫ π π π  06.0cos1 65 t π 6cos65dtt π 65sinidtq 100    Chapter 1, Solution 5 µCmC)e1( 21e21-mCdteidtq 4202t-2t-  490 =−==== ∫ ∫   Chapter 1, Solution 6 (a) At t = 1ms, mA40 === 280 dt dq i   (b) At t = 6ms, mA0 == dt dq i   (c) At t = 10ms, mA20- === 480 dt dq i Chapter 1, Solution 7 <<<<<<== 8t6 25A, 6t2 25A,- 2t0 A,25 dtdqi which is sketched below:
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