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Fig. 1-18 (a) With terminals a; b open-circuited, only loop current I s ows; by KVL and Ohm s law, with all currents and voltages understood to be functions of s, we have I Now KVL gives VTh Vab V1 sLI V1 sL V2 V1 V1 s2 LCV2 sL 1=sC s2 LC 1 V2 V1 sL 1=sC

x) ; (e) 100 ; ( f ) 80 ; (g) 55 ; (h) 72

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The passive circuit elements resistance R, inductance L, and capacitance C are de ned by the manner in which the voltage and current are related for the individual element. For example, if the voltage v and current i for a single element are related by a constant, then the element is a resistance, R is the constant of proportionality, and v Ri. Similarly, if the voltage is the time derivative of the current, then the element is an inductance, L is the constant of proportionality, and v L di=dt. Finally, if the current in the element is the time derivative of the voltage, then the element is a capacitance, C is the constant of proportionality, and i C dv=dt. Table 2-1 summarizes these relationships for the three passive circuit elements. Note the current directions and the corresponding polarity of the voltages.

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19. (a) 85 ; (b) y ; (c) 110 ; (d) 95 ; (e) 72 ; ( f ) 50 ; (g) 145 ; (h) 87

With the independent sources deactivated, the Thevenin impedance can be determined as ZTh sLk (b) The Norton current can be found as V IN Th ZTh and the Norton admittance as YN 1 s2 LC 1 ZTh sL V1 s2 LCV2 2 s2 LC 1 V1 s LCV2 sL sL s2 LC 1 1 sL 1=sC sL sC sL 1=sC s2 LC 1

ohms ( )

1 2 x;

v Ri (Ohms s law)

Determine the z parameters for the two-port network of Fig. 1-19.

1 2 x;

henries (H)

CHAP. 1]

50 y

farads (F)

22. (a) 19; (b) 45; (c) 69; (d) 90; (e) 125; ( f ) 167; (g) 23. (a) 110; (b) 135; (c) 180; (d) 270; (e) 180 24. (a) 45 ; (b) 60 ; (c) 30 ; (d) 18

10 W 3

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All electrical devices that consume energy must have a resistor (also called a resistance) in their circuit model. Inductors and capacitors may store energy but over time return that energy to the source or to another circuit element. Power in the resistor, given by p vi i2 R v2 =R, is always positive as illustrated in Example 2.1 below. Energy is then determined as the integral of the instantaneous power t2 t2 1 t2 2 2 w p dt R i dt v dt R t1 t1 t1

(h) 180

EXAMPLE 2.1. A 4.0- resistor has a current i 2:5 sin !t (A). Find the voltage, power, and energy over one cycle. ! 500 rad/s. v Ri 10:0 sin !t V p vi i2 R 25:0 sin2 !t W ! t t sin 2!t J w p dt 25:0 2 4! 0 The plots of i, p, and w shown in Fig. 2-6 illustrate that p is always positive and that the energy w, although a function of time, is always increasing. This is the energy absorbed by the resistor.

V2 _

2x; ( f ) 360

Fig. 2-6

Fig. 1-19 Also, at node b, KCL gives I1 0:3Ia Ia 1:3Ia 1:3 Thus, by (1.10), z11 Further, again by Ohm s law, Ia Substitution of (2) into (1) yields I1 1:3 so that, by (1.12), z21 V2 6 V2 6 2 V1 16 1

26. (a) 75 ; (b) 75 ; (c) 115 ; (d) 100 ; (e) 140 ; ( f ) 230 ; (g) 80 ; (h) 48 27. (a) 85 ; (b) 103 ; (c) 80 ; (d) 72 ; (e) 90 ; ( f ) 110 ; (g) 130 ; (h) 110 28. (a) m x 68 , mjy 95 ; (b) mjx 90 , mjy 120 ; (c) m x

CHAP. 2]

V1 16 12:308 I1 I2 0 1:3

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