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After the switching, the three inductances have the equivalent Leq Then 5=200 25 ms, and so, with t in ms, i 6e t=25 and A i2 5 i 2e t=25 15 A 10 5 10 5H 6 15

6. Any point on the line which is the bisector of the angle between two lines is equidistant from the sides of the angle.

(a) With the interelectrode capacitances in position, the small-signal equivalent circuit is given by Fig. 7-24. The input admittance is Yin IS I I2 1 VS VS VS sCgk VS 1=sCgk VS Vo sCpg VS Vo 1=sCpg 1

Fig. 7-31 As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence, Req C 5

6. Any point on the plane which is the bisector of the dihedral angle between two planes is equidistant from the sides of the angle.

2 mF 10 ms By continuity, vC 0 vC 0 0. Furthermore, as t ! 1, the capacitor becomes an open circuit, leaving 20

(2) (3)

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Extension of Locus Principles The following dual statements involve the locus of points in a plane and in space. Locus in a Plane 1. The locus of points at a given distance from a given point is a circle having the given point as center and the given distance as radius. Locus in Space 1. The locus of points at a given distance from a given point is a sphere having the given point as center and the given distance as radius.

25 V

Locus in a Plane 2. The locus of points 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.

CHAP. 7]

Knowing the end conditions on vC , we can write vC vC 0 vC 1 e t= vC 1 25 1 e t=10 wherein t is measured in ms. The current in the capacitor is given by iC C and the current in the parallel 10-

Locus in Space 2. The locus of points at a given distance from a given plane is a pair of planes parallel to the given plane and at the given distance from it.

Fig. 7-24

The problem might also have been solved by assigning mesh currents and solving simultaneous di erential equations.

3. The locus of points equidistant from two points is the line that is the perpendicular bisector of the segment joining the two points.

CHAP. 7]

Substituting (2) and (3) into (1) and rearranging give V Yin s Cgk 1 o Cpg VS Now, from the result of Problem 7.21, Vo RL VS RL rp so (4) becomes RL Yin s Cgk 1 C RL rp pg

3. The locus of points equidistant from two points is the plane that is the perpendicular bisector of the segment joining the two points.

The steady-state solution (particular solution) of (26) is i1 1 3333=833 4:0 A; hence i1 Ae 833t 4:0 A The initial condition i1 0 i1 0 0 now gives A 4:0 A, so that i1 4:0 1 e 833t A and i2 4:0 2:67e 833t A

4. The locus of points equidistant from the lines that are the sides of an angle is the line which is the bisector of the angle between them.

(b) The output admittance is Yo and I2 Ip Ipk IL Vo Vo Ipk sCpk Vo 7 (8)

When the rest of the circuit is viewed from the terminals of the inductance, there is equivalent resistance Req 5 5 10 8:33

4. The locus of points equidistant from the planes that are the sides of a dihedral angle is the plane which is the bisector of the angle between them.

Let Yo0 be the output admittance that would exist if the capacitances were negligible; then so that (c) Yo s Ip Yo0 Vo R L rp 1 Cpg Cpk Yo0 RL 9 (10)

Then 1= Req =L 833 s 1 . At t 1, the circuit resistance is RT 10 5 5 12:5

5. The locus of points equidistant from two parallel lines is the line parallel to them and midway between them.

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