HW1_Soln - August 31, 2009 ECE2100 Homework # 1 Solutions...

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Unformatted text preview: August 31, 2009 ECE2100 Homework # 1 Solutions (Rev. 0) Fall 2009 1. Textbook Chapter 1 Problems A. P1.8 The current flow down a conductor of cross sectional area A can be expressed as I = dQ dt =- qnAv n , where q is the charge per electron [1 . 603(10- 19 ) C], n is the electron volume concentra- tion, and v n is the electron velocity down the conductor. The negative sign is to reflect the charge of the electron (and the adopted current convention). In a incremental slice of the conductor dx , the total number of electrons is given as: N n = nAdx . Recognizing that the velocity is dx/dt then combining these relations gives: dN n dt =- I q =- 35(10- 6 ) 1 . 602(10- 19 ) =- 2 . 1848(10 14 ) electrons/sec . B. P1.9 i = dQ dt- Q = Z idt + C = Z 24cos(4000 t ) dt + C = 24 4000 sin(4000 t ) + C , where C is an integration constant. Now at a time of 0 where the current is maximum (cosine function is at its maximum) the charge is zero. Accordingly, Q (0) = 0 = 0 + C- C = 0 . Finally, Q ( t ) = 24 4000 sin(4000 t ) = 0 . 0060sin(4000 t ) [ C ] . C. P1.11 ECE2100 August 31, 2009 Page 2 DC steady state power. E = Pt = V It = 9(0 . 1)5(60)(60) = 16 . 2 kJ . D. P1.12 In case a) 120(5) = 600 Watts of power - flows from A to B (positive voltage & current). In case b) 250(- 8) =- 2000 Watts of power - flows from B to A (positive voltage & negative current). In case c)- 150(16) =- 2400 Watts of power - flows from B to A (negative voltage & positive current). In case d)- 480(- 10) = 4800 Watts of power flows from A to B (negative voltage & current). E. P1.13 In case a) 400 W = 40(10) is being delivered from the box. Terminal 2 has electrons leaving the box (as current is entering). Electrons gain energy as they move to a more negative potential. In case b) 400 W is being delivered to the box. Terminal 2 has electrons entering the box (as current is leaving). Electrons loose energy as they move to a more positive potential. In case c) 400 W is being delivered to the box. Terminal 2 has electrons leaving the box (as current is entering). Electrons loose energy as they move to a more positive potential. In case d) 400 W is being delivered from the box. Terminal 2 has electrons entering the box (as current is leaving). Electrons gain energy as they move to a more negative potential. 2. More Textbook Chapter 1 Problems. A. P1.21 ECE2100 August 31, 2009 Page 3 P ( t ) = v ( t ) i ( t ) = 36(25)sin(200 t )cos(200 t ) = 36(25) 2 sin(400 t ) = 450sin(400 t ) [ W ] ....
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HW1_Soln - August 31, 2009 ECE2100 Homework # 1 Solutions...

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