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The same line of reasoning applies to the second mesh Figure 314 depicts the voltage assignment around the second mesh, following the clockwise direction of mesh current i2 The mesh current i2 is also the branch current through resistors R3 and R4 ; however, the current through the resistor that is shared by the two meshes, R2 , is now equal to (i2 i1 ), and the voltage across this resistor is v2 = (i2 i1 )R2 and the complete expression for mesh 2 is (i2 i1 )R2 + i2 R3 + i2 R4 = 0 (315) (314)

x =0

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vS + _

61 65

Why is the expression for v2 obtained in equation 314 different from equation 312 The reason for this apparent discrepancy is that the voltage assignment for each mesh was dictated by the (clockwise) mesh current Thus, since the mesh currents ow through R2 in opposing directions, the voltage assignments for v2 in the two meshes will also be opposite This is perhaps a potential source of confusion in applying the mesh current method; you should be very careful to carry out the assignment of the voltages around each mesh separately Combining the equations for the two meshes, we obtain the following system of equations: (R1 + R2 )i1 R2 i2 = vS R2 i1 + (R2 + R3 + R4 )i2 = 0 (316)

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During eukaryotic cell division, the processes of mitosis and meiosis apportion the chromosomes to daughter cells. Both processes are preceded by chromosome replication during the S phase of the cell cycle, which is under genetic control. In mitosis, the two sister chromatids making up each replicated chromosome separate into two daughter cells. Sex cells gametes in animals and spores in plants are produced by the two-stage process of meiosis. In meio-

These equations may be solved simultaneously to obtain the desired solution, namely, the mesh currents, i1 and i2 You should verify that knowledge of the mesh currents permits determination of all the other voltages and currents in the circuit The following examples further illustrate some of the details of this method

+ V2 _

The behavior of chromosomes during meiosis explains Mendel s two principles, segregation and independent assortment. At the end of this chapter, we de ne the chromosomal theory of inheritance, the concept that shapes the rst section of this book. This theory states that genes are located on chromosomes; their positions and order on the chromosomes can be discovered by mapping techniques described in later chapters.

Fig 817: Effective mass of a cantilever beam [8]

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+ V3 _

R1 = 5

The McGraw Hill Companies, 2001

Schematics, Diagrams, Circuits, and Given Data: V1 = 10 V; V2 = 9 V; V3 = 1 V;

PROBLEM 1: What are the differences between chromo-

Assumptions: Assume clockwise mesh currents i1 and i2 Figure 315 Analysis: The circuit of Figure 315 will yield two equations in two unknowns, i1 and i2 It is instructive to consider each mesh separately in writing the mesh equations; to this end, Figure 316 depicts the appropriate voltage assignments around the two meshes,

Evaluating at x = l,

based on the assumed directions of the mesh currents From Figure 316, we write the mesh equations: V1 R1 i1 V2 R2 (i1 i2 ) = 0 R2 (i2 i1 ) + V2 R3 i2 V3 R4 i2 = 0 Rearranging the linear system of the equation, we obtain 15i1 10i2 = 1 10i1 + 20i2 = 8 which can be solved to obtain i1 and i2 : i1 = 05 A and i2 = 065 A

that only one chromosome from each homologous pair goes into a gamete; this is also true for the maternal and paternal alleles of a given gene. Independent assortment is explained by the independent behavior of each tetrad at meiosis. That is, the separation of maternal and paternal alleles in one tetrad is independent of the separation of maternal and paternal alleles in any other tetrad.

Comments: Note how the voltage v2 across resistor R2 has different polarity in Figure 316, depending on whether we are working in mesh 1 or mesh 2

R2 = 8

PROBLEM 3: A hypothetical organism has six chromo-

Assumptions: Assume clockwise mesh currents i1 , i2 , and i3 Analysis: Starting from mesh 1 we apply KVL to obtain

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