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The unit of current, the ampere (A), is de ned as the constant current in two parallel conductors of in nite length and negligible cross section, 1 meter apart in vacuum, which produces a force between the conductors of 2:0 10 7 newtons per meter length. A more useful concept, however, is that current results from charges in motion, and 1 ampere is equivalent to 1 coulomb of charge moving across a xed surface in 1 second. Thus, in time-variable functions, i A dq=dt C/s). The derived unit of charge, the coulomb (C), is equivalent to an ampere-second. The moving charges may be positive or negative. Positive ions, moving to the left in a liquid or plasma suggested in Fig. 1-1(a), produce a current i, also directed to the left. If these ions cross the plane surface S at the rate of one coulomb per second, then the resulting current is 1 ampere. Negative ions moving to the right as shown in Fig. 1-1(b) also produce a current directed to the left.



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Table 1-3 Pre x pico nano micro milli centi deci kilo mega giga tera Factor 10 10 9 10 6 10 3 10 2 10 1 103 106 109 1012

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There is evidence that the worldwide demand for industrial robots has yet to reach the numbers predicted by industrial experts and visionaries

Fig. 1-1





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Of more importance in electric circuit analysis is the current in metallic conductors which takes place through the motion of electrons that occupy the outermost shell of the atomic structure. In copper, for example, one electron in the outermost shell is only loosely bound to the central nucleus and moves freely from one atom to the next in the crystal structure. At normal temperatures there is constant, random motion of these electrons. A reasonably accurate picture of conduction in a copper conductor is that approximately 8:5 1028 conduction electrons per cubic meter are free to move. The electron charge is e 1:602 10 19 C, so that for a current of one ampere approximately 6:24 1018 electrons per second would have to pass a xed cross section of the conductor.

EXAMPLE 1.2. A conductor has a constant current of ve amperes. How many electrons pass a xed point on the conductor in one minute 5 A 5 C=s 60 s=min 300 C=min 300 C=min 1:87 1021 electrons=min 1:602 10 19 C=electron

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Fig. 5-27

some twenty years ago The early industrial robots were expensive and temperamental, and they required a lot of maintenance Moreover, the software was frequently inadequate for the assigned tasks, and many industrial robots were ill-suited to the tasks assigned them Many early industrial customers in the 1970s and 1980s were disappointed because their expectations had been unrealistic; they had underestimated the costs involved in operator training, the preparation of applications software, and the integration of the industrial robots with other machines and processes in the workplace By the late 1980s, the decline in orders for industrial robots drove most American companies producing them to go out of business, leaving only a few small, generally unrecognized manufacturers Such industrial giants as General Motors, Cincinnati Milacron, General Electric, International Business Machines, and Westinghouse entered and left the field.

The inverting ampli ers, summing circuits, and integrators described in the previous sections are used as building blocks to form analog computers for solving linear di erential equations. Di erentiators are avoided because of considerable e ect of noise despite its low level. To design a computing circuit, rst rearrange the di erential equation such that the highest existing derivative of the desired variable is on one side of the equation. Add integrators and ampli ers in cascade and in nested loops as shown in the following examples. In this section we use the notations x 0 dx=dt, x 00 d 2 x=dt2 and so on.

EXAMPLE 5.20 Design a circuit with x t as input to generate output y t which satis es the following equation: y 00 t 2y 0 t 3y t x t Step 1. Rearrange the di erential equation (27) as follows: y 00 x 2y 0 3y 28 27

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