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From (20), q 10 1 e mC; and we know that q 1 0:5 mF 40 V 20 mC. Hence, q, is determined for t ! as q q q 1 e t = q 1 71:55e 4000t 20 Di erentiating (20) and (21), dq i dt See Fig. 7-27. 40e 4000t 286:2e 4000t mA mA 0 < t < t > mC 21

Regular Octahedron (8 faces)

(b) The ampli cation factor is most easily evaluated along the vP axis. From (4.9), for the point iP 0, vP 100 V; vG 4 V, we obtain vP 100 25 4 vG

Fig. 7-27

Regular Dodecahedron (12 faces)

Use the current-source small-signal triode model of Fig. 7-9(a) to derive the voltage-source model of Fig. 7-9(b).

The current in an RL circuit is a continuous function, starting at zero in this case, and reaching the nal value V=R. Thus, for t > 0, i V 1 e t= R and vR Ri V 1 e t=

Regular Icosahedron (20 faces)

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We need to nd the Thevenin equivalent for the circuit to the left of the output terminals in Fig. 7-9(a). If the independent source is deactivated, then vg 0; thus, gm vg 0, and the dependent current source acts as an open circuit. The Thevenin resistance is then RTh rp . The open-circuit voltage appearing at the output terminals is vTh gm vg rp vg where gm rp is the ampli cation factor. Proper series arrangement of vTh and RTh gives the circuit of Fig. 7-9(b).

V 1 e t= 1 V 2 e t= 1 2 t ln 2 that is, when t 0:693 . Note that this time is independent of V.

Fig. 17-11

A constant voltage is applied to a series RL circuit at t 0. 20 V at 3.46 ms and 5 V at 25 ms. Obtain R if L 2 H.

For the ampli er of Example 7.6, (a) use (7.9) to evaluate the plate resistance and (7.10) to nd the transconductance.

Regular hexagons cannot be faces of a regular polyhedron. If three regular hexagons had a common vertex (Fig. 17-12), the sum of the measures of the three interior angles at that vertex would be 3(120 ) or 360 . As a result, the three regular hexagons would lie in the same plane, so they could not form a solid.

Using the two-point method of Section 7-6.

Fig. 17-12

a rp % vP 218 152 10 k iP vG 4 14:7 8:1 10 3

CHAP. 7]

Sphere in space geometry corresponds to circle in plane geometry. Similarly, plane corresponds to straight line. By interchanging circle and sphere, or straight line and plane, each of the following dual statements can be interchanged. When you do so, many plane geometry principles with which you are acquainted become space geometry principles. Related Dual Statements 1. Every point on a circle is at a distance of one radius from its center. 2. Two intersecting straight lines determine a plane.

gm %

t2 t1 25 3:46 15:54 ms ln v1 ln v2 ln 20 ln 5 R L 2 128:7

1. Every point on a sphere is at a distance of one radius from its center. 2. Two intersecting planes determine a straight line.

15:54 10 3

Find an expression for the voltage gain Av vp =vg of the basic triode ampli er of Fig. 4-12, using an ac equivalent circuit.

Be sure, in obtaining dual statements, that there is a complete interchange of terms. If the interchange is incomplete, as in the following pair of statements, there is no duality. 4. The locus of a point at a given distance from a given line is a pair of lines parallel to the given line and at the given distance from it. 4. The locus of a point at a given distance from a given line is a circular cylindrical surface having the given line as axis and the given distance as radius.

Fig. 7-28 As there is always inductance in the circuit, the current is a continuous function at all times. In the interval 0 t 4 ms, with the 100

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