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4 = 200 j 4,468 LC
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These are complex conjugate roots, therefore we have an underdamped circuit with natural response given by equation 565 By analogy with Example 513, the complete solution is given by: iL (t) = iLN (t) = K1 e( 200+j 4,468)(t
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+ K2 e( 200 j 4,468)(t
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The initial conditions for the energy storage elements are: vC ( t ) = 0 V; iL ( t + ) = 596 A Thus, iL ( t + ) = 596 = K1 e0 + K2 e0 K1 + K2 = 596 K1 = 596 K2 To use the second initial condition, we observe that 1 diL vL ( t + ) ( t +) = dt LP and we note that the inductor voltage can change instantaneously; ie, vL (0 ) = vL (0+ ) To determine vL (0+ ) we need to apply KVL once again at t = 0+ : V B vC ( t + ) v R ( t + ) v L ( t + ) = 0 vR ( t + ) = iL ( t + )R = 596 2 = 1192 vC ( t + ) = 0 Therefore vL ( t + ) = VB 1192 = 12 1192 = 008 and vL ( t + ) 008 diL ( t +) = = = 16 dt LP 5 10 3 Now we can obtain a second equation in K1 and K2 , diL ( t + ) = ( 200 + j 4, 468)K1 e0 + ( 200 j 4, 468)K2 e0 dt and since K1 = 596 K2 16 = ( 200 + j 4, 468) (596 K2 ) + ( 200 j 4, 468)K2 = 1192 + j 26, 629 ( 200 + j 4, 468)K2 + ( 200 j 4, 468)K2 = 1192 + j 26, 629 j 8936K2 K2 = 1 (1208 j 26,629) = 298 + j 01352 j 8,936 V
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K1 = 596 K2 = 298 j 01352 Note, again, that K1 and K2 are complex conjugates, as suggested earlier
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Finally, the complete solution is: iL (t) = (298 j 01352)e( 200+j 4,468)(t = 298e
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The coil primary current is plotted in Figure 553
6 4 2 Amperes 0 2 4 6 0 0005 001 0015 002 0025 Time (s) 003 0035 004
Figure 553 Ignition circuit primary current response
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
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