birt code 39

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To compute the primary voltage, we simply differentiate the inductor current and multiply by LP ; to determine the secondary voltage, which is that applied to the spark plug, we simply remark that a 1:100 transformer steps up the voltage by a factor of 100, so that the secondary voltage is 100 times larger than the primary voltage Thus, the expression for the secondary voltage is: d diL vspark plug = 100 LP = 05 (2 298e 200t cos(4,468t) dt dt 2 01352e 200t sin(4,468t) = 05 [ 200 596 e 200t cos(4,468t) 4,468 596 e 200t sin(4,468t)] 05 [ 200 01352 e 200t sin(4,468t) + 4,468 01352 e 200t cos(4,468t)] where we have reset time to t = 0 for simplicity We are actually interested in the value of this voltage at t = 0, since this is what will generate the spark; evaluating the above expression at t = 0, we obtain: vspark plug (t = 0) = 05 [ 200 596] 05 [4,468 01352] vspark plug (t = 0) = 596 302 = 898 V One can clearly see that the result of the switching is a very large (negative) voltage spike, capable of generating a spark across the plug gap A plot of the secondary voltage starting at the time when the switch is opened is shown in Figure 554, showing that approximately 03 ms after the switching transient, the secondary voltage reaches approximately 12,500 volts! This value is typical of the voltages required to generate a spark across an automotive spark plug

5

15 1 05 104 volts 0 05 1 15 0 0002 0006 001 0014 Time (s) 0018 002

84 39 45 30 58 53

Figure 554 Secondary ignition voltage response Focus on Computer-Aided Tools: The Matlab m- le containing the numerical analysis and plotting commands for this example may be found in the CD that accompanies this book An EWB simulation is also enclosed

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59 Derive the differential equation for the series circuit of Figure 539(b) Show that one can write the equation either as RT dvC 1 1 d 2 vC + + vC = vT (t) dt 2 L dt LC LC or as 1 RT diL 1 dvT (t) d 2 iL + iL = + 2 dt L dt LC L dt 510 Determine the roots of the characteristic equation of the series RLC circuit of Figure 539(b) with R = 100 , C = 10 F, and L = 1 H 511 For the series RLC circuit of Figure 539(b), with L = 1 H and C = 10 F, nd the ranges of values of R for which the circuit response is overdamped and underdamped, respectively

Length Width lblArea (Leave this blank) txtLength (Leave this blank) txtWidth (Leave this blank) btnCalculate Click to Calculate Area

CYU 51 CYU 52 CYU 53 CYU 54 CYU 55 CYU 56 CYU 57 CYU 58 CYU 59 CYU 510 1 dvC 1 + vC = vS dt RC RC 1 1 dv + v = iS dt RC C R diL R + iL = iS dt L L vC (0 ) = 8 V and iR (0+ ) = 2 A vC = 75 75e t/0025 V, 0 t < 005 s; vC = 6485e (t 005)/00375 V, t 005 s t90% = 125 s 545 s The output pulse has a higher peak 50 j 31225 Overdamped: R > 63246 ; underdamped: R < 63246

through the inductor is 170 mA in the direction shown in Figure P51 Did steady-state conditions exist just before the switch was opened L = 09 mH R1 = 6 k R3 = 3 k VS = 12 V R2 = 6 k

+ V _ S _

131 51 73 58 96 71

R2 iL R1 L R3 + VR3 _

P51, the current through the inductor is 170 mA in the direction shown Determine the voltage across R3 just after the switch is opened VS = 12 V R1 = 6 k R3 = 3 k L = 09 mH R2 = 6 k

In addition, change the text property of the form to read Web Service Test Bed. Arrange the six controls according to Figure 10-5.

and just after the switch is changed in Figure P56 Assume steady-state conditions exist for t < 0 VS = 12 V R1 = 22 k Rs = 07 L = 100 mH

C + + VS 1 _

Figure P57 at t < 0 The switch is closed at t = 0 V1 = 12 V R2 = 22 k C = 047 F R1 = 068 k R3 = 18 k

before and just after the switch is closed in Figure P53 Assume steady-state conditions for t < 0 V1 = 12 V R1 = 068 k C = 05 F R2 = 18 k

151 79 93 90 133 105

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Figure 10-5. The Design form for the Web Service test bed 3. On the Project menu, click the Add Web Reference submenu item. The dialog box in Figure 10-6 appears.

ic(t) R2 C R3 t=0 R1 + + V 1 _

R2 Io R1 C + VC(t)

circuit shown in Figure P514 for t < 0 The switch is changed at t = 0 as shown VS1 = 17 V R1 = 14 k R3 = 14 k VS2 = 11 V R2 = 13 k C = 70 nF

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