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45. Let us expand the determinant of the coe cients in terms of the elements of the rst row; doing this, you should nd that    4 18 7      D  2 4p p  46p2 92p 138   p 3 5 The system will have non-trivial solutions only if D 0. Thus, setting D 0, we have the requirement that 46p2 92p 138 0, the solutions of which are (using the quadratic formula),* p 1 or p 3, answers to rst part. Next, setting rst p 1 and then p 3 in the original given system gives the two possible systems 4x 18y 7z 0 2x 4y z 0 x 3y 5z 0 4x 18y 7z 0 2x 12y 3z 0 3x 3y 5z 0
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Now write down the values of a through i for each of the above two equations, then substitute into eq. (55). Doing this, you should nd that the answers are for p 1 : x 23; y 9; z 10; thus x 23k thus y 9k thus z 10k for p 3 : x 69; y 1; z 42; thus x 69k thus y k thus z 42k
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46. In Fig. 48 the current arrows show the direction of ow of what we will call positive current. Let us therefore write the voltage equation generated by tracing around the circuit in the cw sense, putting voltage drops on the left-hand side and voltage rises on the right-hand side, as agreed upon in the discussion of Fig. 47. Doing this, starting at point A (keeping Fig. 46 in mind), we have that the voltage equation for Fig. 48 is 12 4I 2I I 26 5I 17 thus, I 21=12 1:75 amperes; answer: The answer, minus 1.75, means that, for the given battery voltages and polarities, the current I would actually ow in the ccw sense. 47.{ Let us follow the suggested three steps (as illustrated in the example problem) thus: Step I. Note that two loop currents will satisfy the requirement that all the circuit elements be traversed at least once by a current. Call the current in the left-hand loop I1 and the current in the right-hand loop I2 , and let us elect to draw the arrow-heads to indicate that I1 and I2 both ow in the clockwise sense.
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* From note 1 in the Appendix. { Unless speci cally stated otherwise, we ll assume that all generators (batteries in these problems) have negligible internal resistance. (See Figs. 25 and 26 in section 2.5 for a discussion of internal resistance. )
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Step II. voltage equation around left-hand loop: voltage equation around right-hand loop: 15I1 7I2 6 7I1 11I2 21
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Step III. The current in the 7-ohm resistance is the algebraic sum of I1 and I2 , and therefore we must nd the values of both I1 and I2 , as follows. First,    15 7   D   7 11  116 Then,   6   21     15 7  6     7 21  11  1:83621 amperes and I2 3:07759 amperes: 116 116
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For the directions we have elected to take for I1 and I2 , we have the condition shown in the gure to the right (showing just that part of the circuit we re interested in now). Since the two currents in the 7-ohm resistance ow in opposite directions through the resistance, we must take the di erence of the two currents; thus the current in the 7-ohm resistance is I1 I2 1:24138 amperes; answer I2 I1 1:24138 amperes; answer: Both answers mean the same thing; the rst answer means the resultant current (in the 7-ohm resistor) is owing opposite to I1 , that is, in the direction of I2 , while the second answer directly states that the resultant current is owing in the direction of I2 . An ammeter placed in series with the 7-ohm resistor would read 1.24138 amperes. 48. Step I. Three loop currents must be used, to satisfy the requirement that all the circuit elements must be included in the analysis. Let us suppose we elect to have all the currents ow in the clockwise sense, as in the gure below.
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Step II. voltage equation for loop 1: voltage equation for loop 2: voltage equation for loop 3: