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Congruent figures are figures that have the same size and the same shape; they are the exact duplicates of each other. Such figures can be moved on top of one another so that their corresponding parts line up exactly. For example, two circles having the same radius are congruent circles. Congruent triangles are triangles that have the same size and the same shape. If two triangles are congruent, their corresponding sides and angles must be congruent. Thus, congruent triangles ABC and ArBrCr in Fig. 3-1 have congruent corresponding sides (AB ArCr , BC BrCr , and /Cr). AC ArCr ) and congruent corresponding angles (/A /Ar, /B /Br, and /C

EXAMPLE 2.3. In the interval 0 > t > 5 ms, a 20-mF capacitance has a voltage v 50:0 sin 200t (V). Obtain the charge, power, and energy. Plot wC assuming w 0 at t 0. q Cv 1000 sin 200t mC dv 0:20 cos 200t A dt p vi 5:0 sin 400t W t2 p dt 12:5 1 cos 400t mJ wC i C

Voltage division then gives Ie RE RE sCC I V hib RE S hib RE sCC RE hib = RE hib 1 S 1

Fig. 3-1

In the interval 0 > t > 2:5 ms the voltage and charge increase from zero to 50.0 V and 1000 mC, respectively. Figure 2-8 shows that the stored energy increases to a value of 25 mJ, after which it returns to zero as the energy is returned to the source.

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(Read nABC nArBrCr as Triangle ABC is congruent to triangle A-prime, B-prime, C-prime. ) Note in the congruent triangles how corresponding equal parts may be located. Corresponding sides lie opposite congruent angles, and corresponding angles lie opposite congruent sides.

By current division at the output, VL RL IL RL shfb RL RC CC Ie RC h I RC 1=sCC RL fb e sCC RL RC 1 2

Fig. 2-8

PRINCIPLE 1:

Every circuit diagram can be constructed in a variety of ways which may look di erent but are in fact identical. The diagram presented in a problem may not suggest the best of several methods of solution. Consequently, a diagram should be examined before a solution is started and redrawn if necessary to show more clearly how the elements are interconnected. An extreme example is illustrated in Fig. 2-9, where the three circuits are actually identical. In Fig. 2-9(a) the three junctions labeled A

The two coupling capacitors in the CB ampli er of Fig. 6-15 are identical. Also, hrb hob 0. (a) Find an expression for the current-gain ratio Ai s that is valid at any frequency. (b) Find an expression for the midfrequency current-gain ratio.

If two triangles are congruent, then their corresponding parts are congruent. (Corresponding parts of congruent triangles are congruent.)

CHAP. 2]

(a) The low-frequency equivalent circuit is displayed in Fig. 8-18. RE I Ie hib RE S and IL shfb RL RC CC Ie RC h I RC 1=sCC RL fb e sCC RL RC 1 By current division, 1

Fig. 3-2

v vR vC V sin !t q and tan 1 1=!CR V I R2 1=!C 2

PRINCIPLE 2:

Substituting (1) into (2) and dividing by IS give the desired current-gain ratio: Ai s shfb RL RC RE CC IL Ii hib RE sCC RL RC 1 3

The negative phase angle shifts v to the right of the current i. Consequently i leads v for a series RC circuit. The phase angle is constrained to the range 08 908. For 1=!C ( R, the angle 08, and for 1=!C ) R, the angle 908. See Fig. 9-4.

(Side-Angle-Side, SAS) If two sides and the included angle of one triangle are congruent to the corresponding parts of another, then the triangles are congruent.

Fig. 9-4

(b) The midfrequency current-gain ratio is found by letting s ! 1 in (3): Ai 1 hfb RL RC RE hib RE RL RC 4

A brief look at the voltage and current sinusoids in the preceding examples shows that the amplitudes and phase di erences are the two principal concerns. A directed line segment, or phasor, such as that shown rotating in a counterclockwise direction at a constant angular velocity ! (rad/s) in Fig. 9-5, has a projection on the horizontal which is a cosine function. The length of the phasor or its magnitude is the amplitude or maximum value of the cosine function. The angle between two positions of the phasor is the phase di erence between the corresponding points on the cosine function.

Fig. 3-3

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