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Now, with RTh as found in Problem 1.7, YN 1 1 0:4167 A RTh 2:4

In general, a coil can be represented by either a series or a parallel arrangement of circuit elements. The frequency of the applied voltage may require that one or the other be used to represent the device.

6

A voltage function and a polarity must be speci ed to completely describe a voltage source. The polarity marks, and , are placed near the conductors of the symbol that identi es the voltage source. If, for example, v 10:0 sin !t in Fig. 2-4(a), terminal A is positive with respect to B for 0 > !t > , and B is positive with respect to A for > !t > 2 for the rst cycle of the sine function.

For the circuit and values of Problems 1.7 and 1.8, nd the Thevenin impedance as the ratio of open-circuit voltage to short-circuit current to illustrate the equivalence of the results.

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5. (a) square; (b) isosceles triangle; (c) trapezoid; (d) right triangle 6. (a) 140 ; (b) 60 ; (c) 90 ; (d) (180 8. (a) x 22; (b) y 6; (c) AB CD x) ; (e) x ; ( f ) (90 22; (d) perimeter x) 44; (e) x 21; ( f ) r 14 7. (a) 100 ; (b) 50 , 80 ; (c) 54 , 27 ; (d) 45 ; (e) 35 ; ( f ) 45 9. (a) 0; (b) 40; (c) 33; (d) 7 10. (a) tangent externally; (b) tangent internally; (c) the circles are 5 units apart; (d) overlapping 11. (a) concentric; (b) tangent internally; (c) tangent externally; (d) outside each other; (e) the smaller entirely inside the larger; ( f ) overlapping 13. (a) 40; (b) 90; (c) 170; (d) 180; (e) 2x; ( f ) 180 15. (a) 85; (b) 170; (c) c; (d) 2i; (e) 60; ( f ) 30 16. (a) 60, 120, 180; (b) 80, 120, 160; (c) 100, 120, 140; (d) 36, 144, 180 17. (a) mjx 136 ; (b) m y x; (g) 2x 2y

Fig. 2-4

1 2 x;

The open-circuit voltage is VTh as found in Problem 1.7, and the short-circuit current is IN from Problem 1.8. Thus, ZTh which checks with the result of Problem 1.7. VTh 12 2:4 5 IN

Similarly, a current source requires that a direction be indicated, as well as the function, as shown in Fig. 2-4(b). For passive circuit elements R, L, and C, shown in Fig. 2-4(c), the terminal where the current enters is generally treated as positive with respect to the terminal where the current leaves. The sign on power is illustrated by the dc circuit of Fig. 2-5(a) with constant voltage sources VA 20:0 V and VB 5:0 V and a single 5- resistor. The resulting current of 3.0 A is in the clockwise direction. Considering now Fig. 2-5(b), power is absorbed by an element when the current enters the element at the positive terminal. Power, computed by VI or I 2 R, is therefore absorbed by both the resistor and the VB source, 45.0 W and 15 W respectively. Since the current enters VA at the negative terminal, this element is the power source for the circuit. P VI 60:0 W con rms that the power absorbed by the resistor and the source VB is provided by the source VA .

14. (a) 20; (b) 45; (c) 85; (d) 90; (e) 130; ( f ) 174; (g) x; (h) 90

Thevenin s and Norton s theorems are applicable to other than dc steady-state circuits. For the frequency-domain circuit of Fig. 1-18 (where s is frequency), nd (a) the Thevenin equivalent and (b) the Norton equivalent of the circuit to the right of terminals a; b.

CHAP. 2]

11 ; (c) mjx x) ; (d) (90

Fig. 2-5

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