# B the four digit conclusion is quite different 3454

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(b)The four-digit conclusion is quite different.-(.3454/.0003)ρ1+ρ2-→.00031.5561.56901789-1805=x= 10460, y=-1.0094(a)For the first one, first, (2/3)-(1/3) is.666 666 67-.333 333 33 =.333 333 34 and so (2/3) +((2/3)-(1/3)) =.666 666 67 +.333 333 34 = 1.000 000 0. For the other one, first ((2/3) + (2/3)) =.666 666 67 +.666 666 67 = 1.333 333 3 and so ((2/3) + (2/3))-(1/3) = 1.333 333 3-.333 333 33 =.999 999 97.(b)The first equation is.333 333 33·x+ 1.000 000 0·y= 0 while the second is.666 666 67·x+2.000 000 0·y= 0.5(a)This calculation-(2/3)ρ1+ρ2-→-(1/3)ρ1+ρ332160-(4/3) + 2ε-(2/3) + 2ε-2 + 4ε0-(2/3) + 2ε-(1/3)-ε-1 +ε-(1/2)ρ2+ρ3-→32160-(4/3) + 2ε-(2/3) + 2ε-2 + 4ε0ε-2ε-εgives a third equation ofy-2z=-1. Substituting into the second equation gives ((-10/3)+6ε)·z=(-10/3) + 6εsoz= 1 and thusy= 1. With those, the first equation says thatx= 1.(b)The solution with two digits kept.30×101.20×101.10×101.60×101.10×101.20×10-3.20×10-3.20×101.30×101.20×10-3-.10×10-3.10×101-(2/3)ρ1+ρ2-→-(1/3)ρ1+ρ3.30×101.20×101.10×101.60×1010-.13×101-.67×100-.20×1010-.67×100-.33×100-.10×101-(.67/1.3)ρ2+ρ3-→.30×101.20×101.10×101.60×1010-.13×101-.67×100-.20×10100.15×10-2.31×10-2comes out to bez= 2.1,y= 2.6, andx=-.43.
246Linear Algebra, by HefferonTopic: Analyzing Networks1(a)The total resistance is 7 ohms. With a 9 volt potential, the flow will be 9/7 amperes. Inciden-tally, the voltage drops will then be: 27/7 volts across the 3 ohm resistor, and 18/7 volts across eachof the two 2 ohm resistors.(b)One way to do this network is to note that the 2 ohm resistor on the left has a voltage dropacross it of 9 volts (and hence the flow through it is 9/2 amperes), and the remaining portion onthe right also has a voltage drop of 9 volts, and so is analyzed as in the prior item.We can also use linear systems.-→i0i1-→i2i3←-Using the variables from the diagram we get a linear systemi0-i1-i2= 0i1+i2-i3= 02i1= 97i2= 9which yields the unique solutioni1= 81/14,i1= 9/2,i2= 9/7, andi3= 81/14.Of course, the first and second paragraphs yield the same answer. Esentially, in the first para-graph we solved the linear system by a method less systematic than Gauss’ method, solving for someof the variables and then substituting.(c)Using these variables-→i0i1-→i2-→i3i4i5←-i6←-one linear system that suffices to yield a unique solution is this.i0-i1-i2= 0i2-i3-i4= 0i3+i4-i5= 0i1+i5-i6= 03i1= 93i2+ 2i4+ 2i5= 93i2+ 9i3+ 2i5= 9(The last three equations come from the circuit involvingi0-i1-i6, the circuit involvingi0-i2-i4-i5-i6, and the circuit withi0-i2-i3-i5-i6.)Octave givesi0= 4.35616,i1= 3.00000,i2= 1.35616,i3= 0.24658,i4= 1.10959,i5= 1.35616,i6= 4.35616.2(a)Using the variables from the earlier analysis,i0-i1-i2=0-i0+i1+i2=05i1= 208i2= 20-5i1+ 8i2=0The current flowing in each branch is then isi2= 20/8 = 2.5,i1= 20/5 = 4, andi0= 13/2 = 6.5, allin amperes. Thus the parallel portion is acting like a single resistor of size 20/(13/2)3.08 ohms.(b)A similar analysis gives that isi2=i1= 20/8 = 4 andi0= 40/8 = 5amperes. The equivalentresistance is 20/5 = 4 ohms.
Answers to Exercises247(c)Another analysis like the prior ones gives isi2= 20/r2,i1= 20/r1, andi0= 20(r1+r2)/(r1r2), allin amperes. So the parallel portion is acting like a single resistor of size 20/i1=r1r2/(r1+r2) ohms.

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