ACCOUNTING
Solution Chapter 1.docx

B what is the power at the time found in part a c at

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b) What is the power at the time found in part (a)?
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c) At what instant of time is the power being extracted from the circuit element the maximum? d) What is the power at the time found in part (c)? e) Calculate the net energy delivered to the circuit at 0,10, 20, 30, 40 and 50 s. a- p=vi= t(1-.03t) (4-.3t)=(4t-.42t^2+.009t^3) w dp/dt=4-.84t+.027t^2=0 t2=25.24s t1=5.86s p(t1)=10.828 W p(t2)= -21.89W maximum power at t1=5.86s b- pmax=10.828 W(absorbed or delivered) c- maximum negative power (minimum) t2=25.24 d- pmax=21.89 W (supply or extracting) e- w= 0 t pdt = 0 t ( 4 t .42 t 2 + .009 t 3 ) dt = 2 t 2 .14 t 3 + .00225 t 4 w(0)=0 w(10)=82.5 J w(20)=40 J w(40)=0 J w(50)= 1562 J
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9) The voltage and current at the terminals of the circuit element in Fig.(1) are v = 260 cos 850πt V, i = 9 sin 850πt A. a) Find the maximum value of the power being delivered to the element. b) Find the maximum value of the power being extracted from the element. c) Find the average value of p in the interval 0 ≤ t ≤ 3 ms. d) Find the average value of p in the interval 0 ≤ t ≤ 16.525 ms. a- P=vi= 2340cos(850πt) sin(850πt)=1170sin(1700 πt) pmax=1170 w b- pmax(extracting)=1170 w c- pavg= ( 5.1 π ) 1 cos ¿ ¿ 1170sin ( 1700 π t ) dt = 39 10 4 | cos ( 1700 π t ) 1700 π | 3 10 3 0 = 3900 17 π ¿ 1 3 10 3 0 3 10 3 ¿ d- pav= ( 28.0925 π ) 1 cos ¿ ¿ 1170sin ( 1700 π t ) dt = 70.25 10 3 | cos ( 1700 π t ) 1700 π | 16.525 10 3 0 = 41.32 π ¿ 1 16.525
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  • Spring '16
  • Nesma
  • Volt, circuit element, Electrical Engineering course, Prof. Sabah Mohamed, E-JUST University

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