Page 199 - Schaum's Outline of Theory and Problems of Electric Circuits

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
188

Fig. 8-40 [CHAP. 8

8.41 A series RLC circuit contains R ¼ 1
, L ¼ 2 H, and C ¼ 0:25 F. Simultaneously apply magnitude and
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frequency scaling, with K m ¼ 2000 and K f ¼ 10 . What are the scaled element values?
Ans: 2000
; 0:4H; 12:5 mF
8.42 At a certain frequency ! 1 , a voltage V 1 ¼ 25 08 V applied to a passive network results in a current
I 1 ¼ 3:85 308 (A). The network elements are magnitude-scaled with K m ¼ 10. Obtain the current
which results from a second voltage source, V 2 ¼ 10 458 V, replacing the first, if the second source fre-
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quency is ! 2 ¼ 10 ! 1 . Ans: 0:154 158 A
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8.43 In the circuit of Fig. 8-41 let R 1 C 1 ¼ R 2 C 2 ¼ 10 . Find v 2 for t > 0 if: (a) v 1 ¼ cos ð1000tÞuðtÞ,
(b) v 1 ¼ sin ð1000tÞuðtÞ. Ans: ðaÞ v 2 ¼ sin ð1000tÞ; ðbÞ v 2 ¼ 1 cos ð1000tÞ

8.44 In the circuit of Fig. 8-42 assume R ¼ 2 k
, C ¼ 10 nF, and R 2 ¼ R 1 and v 1 ¼ cos !t. Find v 2 for the
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following frequencies: (a) ! 0 ¼ 5 10 rad/s, (b) ! 1 ¼ 10 rad/s.
Ans: ðaÞ v 2 ¼ 2 sin ! 0 t; ðbÞ v 2 ¼ 0:555 cos ð! 1 t 146:38Þ

Fig. 8-41 Fig. 8-42

8.45 Noninverting integrators. In the circuits of Fig. 8-43(a) and 8-43(b) find the relationship between v 2 and v 1 .
Ans: ðaÞ v 1 ¼ðRC=2Þdv 2 =dt; ðbÞ v 1 ¼ 2RCdv 2 =dt

8.46 In the circuit of Fig. 8-44 find the relationship between v 2 and v 1 . Show that for R 1 C 1 ¼ R 2 C 2 we obtain
v 2 ¼ R 2 v 1 =ðR 1 þ R 2 Þ.

dv 2 dv 1
Ans: R 1 R 2 ðC 1 þ C 2 Þ þðR 1 þ R 2 Þv 2 ¼ R 1 R 2 C 1 þ R 2 v 1
dt dt