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Then let n = x in d above: x x < x + 1 so x = x , and x 1 < x x so x = x f Assume that x Z (ie, x is not an integer) Let u = x x and v = x x Then, by c, u 0 and v 0 Also by c, x 1 < x = x u and v + x = x < x + 1, so u < 1 and v < 1 Thus 0 u < 1 and 0 v < 1 But u and v cannot be integers because if either were, then so would x because x = x + u = x v Therefore, 0 < u < 1 and 0 < v < 1, so x = x + u > x and x = x v < x g.

Fig. 14-40

Fig. 14-41 14.32 In the coupled circuit shown in Fig. 14-43, nd V2 for which I1 0. What voltage appears at the 8 inductive reactance under this condition Ans: 141:4 458 V; 100 08 V (+ at dot)

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Let n = x Then ( n) = x so by c, ( x) 1 < ( n) ( x), so x n < x + 1, so x n and n 1 < x, so n 1 < x n Thus by d, n = x Thus x = x , so x = x The second identity follows from the first by replacing x with x h Let n = x + 1 Then by c, (x + 1) 1 < n (x + 1), so x 1 < n 1 x and x = (x + 1) 1 < n Thus, n 1 x < n; that is, (n 1) x < (n 1) + 1 Thus by d, (n 1) = x , so x + 1 = n = x + 1 The proof of the second identity is similar A.

Find the mutual reactance XM for the coupled circuit of Fig. 14-44, if the average power in the 5- resistor is 45.24 W. Ans: 4

n n 1 2 n n 1 n 2 3 un un un n 2! 3!

2 Proof of Theorem A2 on page 320: a Let x = b y Then by definition, log b(b y) = log b(x) = y b Let y = log b x Then by definition, b log b x = b y = x c Let y = log b u and z = log b v Then by definition, u = b y and v = b z, so uv = (b y)(b z) = b y+z, so log b(uv) = y + z = log b u + log b v d By Law c above, log b v + log b u/v = log b (v u/v) = log b u, so log b u/v = log b u + log b u e Let y = log b u Then by definition, u = b y, so uv = (b y )v = b vy.

Fig. 14-42

Fig. 14-43

Then by definition, log b (uv ) = v y = v log b u f Let y = log b x Then by definition, x = b y, so log c x = log c (b y ) = y log c b, by Law e above Thus log b x = y = (log c x)/(log c b) a b c d e f g h i j k l True True True False True True False False False True False True + ar , so S rS.

Fig. 14-44 14.34 For the coupled circuit shown in Fig. 14-45, nd the components of the current I2 resulting from each source V1 and V2 . Ans: 0:77 112:68 A; 1:72 86:058 A

n n 1 n 2 3 6 n2 n un or 0 < u3 < : n 3! n n 1 n 2 lim un 0: Thus lim 1 n n2 1=n lim 1 un 1:

Fig. 14-45 14.35 Determine the coupling coe cient k in the circuit shown in Fig. 14-46, if the power in the 10- resistor is 32 W. Ans: 0:791 In (14a), replace a, X11 , X22 , and XM by their expressions in terms of X1 , X2 , and k, thereby obtaining (14b).

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