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Unlike a cylinder, a cylindrical surface is not limited in extent; nor does it have bases. Similarly, a conical surface is unlimited in the extent and has no base. Extension of Distance Principles Here are some dual statements involving distance in a plane and in space. Distance in a Plane Distance in Space

The equivalent circuit of Fig. 7-9(b) is applicable if RL is connected from P to K. division in the plate circuit, vp RL vg R L rp so Av vp RL vg RL rp Then, by voltage

i 2 1 e t=2 wherein t is measured in ms. In particular,

1. The distance from a point to a line is the length 1. The distance from a point to a plane is the length of the perpendicular from the point to the line. of the perpendicular from the point to the plane.

i 4 2 1 e 2 1:729 A In the interval t ! 4 ms, i starts at 1.729 A and decays toward 100=150 0:667 A, with a time constant 0:1=150 2 ms. Therefore, with t again in ms, 3 i 1:729 0:667 e t 4 = 2=3 0:667 428:4e 3t=2 0:667 A t ! 4 23

In the ampli er of Problem 4.27, let vS 2 cos !t V. (a) Draw the ac load line on Fig. 4-31. (b) Graphically determine the voltage gain. (c) Calculate the voltage gain using small-signal analysis.

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3. The distance from a point to a circle is the external segment of the secant from the point through the center of the circle.

In the circuit of Fig. 7-29, the switch is closed at t 0, when the 6-mF capacitor has charge Q0 300 mC. Obtain the expression for the transient voltage vR .

(a) If capacitor CK appears as a short circuit to ac signals, then application of KVL around the plate circuit of Fig. 4-30 gives, as the equation of the ac load line, VPP VGQ iP RL vP . Thus, the ac load line has vertical and horizontal intercepts VPP VGQ 300 4 25:5 mA RL 11:6 103 as shown on Fig. 4-31. (b) We have vg vS ; thus, as vg swings 2 V along the ac load line from the Q point in Fig. 4-31, vp swings a total of 2Vpm 213 145 68 V as shown. The voltage gain is then Av 2Vpm 68 17 2Vgm 4 and VPP VGQ 296 V

3. The distance from a point to a sphere is the external segment of the secant from the point through the center of the sphere.

The two parallel capacitors have an equivalent capacitance of 3 mF. Then this capacitance is in series with the 6 mF, so that the overall equivalent capacitance is 2 mF. Thus, RCeq 40 ms. At t 0 , KVL gives vR 300=6 50 V; and, as t ! 1, vR ! 0 (since i ! 0). Therefore, vR 50 e t= 50e t=40 in which t is measured in ms. V

5. Any point on the line which is the perpendicular bisector of a segment is equidistant from the ends of the segment.

where the minus sign is included to account for the phase reversal between vp and vg . (c) Applying (7.9) and (7.10) at the Q point of Fig. 4-31 yields vP 202 168 4:86 k rp iP vG 4 15 8 10 3 i 15:5 6:5 10 3 4:5 mS gm P vG vP 180 3 5 Then, gm rp 21:87, and Problem 7.21, yields Av RL 21:87 11:6 103 15:41 R L rp 11:6 4:86 103

Fig. 7-29

5. Any point on the plane which is the perpendicular bisector of a segment is equidistant from the ends of the segment.

Fig. 7-30

The input admittance to a triode modeled by the small-signal equivalent circuit of Fig. 7-9(b) is obviously zero; however, there are interelectrode capacitances that must be considered for highfrequency operation. Add these interelectrode capacitances (grid-cathode capacitance Cgk ; plategrid, Cpg ; and plate-cathode, Cpk ) to the small-signal equivalent circuit of Fig. 7-9(b). Then (a) nd the input admittance Yin , (b) nd the output admittance Yo , and (c) develop a highfrequency model for the triode.

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