Appendix: Answers to Exercises
pffiffiffiffiffi 1 1 1. Chapter 1 Exercise 2.4. FðÞ¼x x kf þ ; 1 0 x x x pffiffiffiffiffi 1 1 1 1 0 SI x kf ¼ V. 0 x x x Exercise 1.1. Physical law. 1 1 0 x2v x2v Exercise 1.2. (a), (b), (f), (g), (i), (j) are quantities. (c), Exercise 2.5. d ¼ a ab ;SI a ab ¼ m. (d), (e), (n) are units. (k) is neither. (l), (m) could be kq xavab kq xavab Exercise 2.6. i2 ¼ 10 mA. quantities if we could devise suitable definitions and = Exercise 2.7. w1 ¼ w0r1 r0. measurements. (h) could be a quantity if we define Exercise 2.8. 14.4 nm. color in terms of frequency. Exercise 2.9. 1 kO. Exercise 1.3. (a) WR (b) J (c) A (d) W (e) s (f) Nm ¼ J. Exercise 2.10. (a) 9:5 O (b) 127 O. 1 t0 0 0 Exercise 1.4. SI L 1 vðt Þdt ¼ SI½ ¼i A. Exercise 2.11. 49:9kO from the E96 series. Exercise 1.5. (a) No, (b) Yes. Exercise 2.12. (a) 4:3kO Æ 5% (b) 459 kO Æ 0:1%. Exercise 1.6. (a) 10 4 As 1, (b) 104 Vs 1, (c) 105 VA, Exercise 2.13. Various answers (d) 1010ms 2. m Exercise 2.14. Because 298 K ¼ 25 C, 298 K is the 1 5 V 5nV m Exercise 1.7. (a) 5mVs ¼ ms ¼ ms , (b) 25kA s¼ reference temperature. 1 1 1 25Ams¼25MAns, (c) 100 mJ ms ¼ 100mJms ¼ Exercise 2.15. a10 ¼ 0:0042 K . 100 Js 1. Exercise 2.16. 4. Exercise 1.8. (a) No space, milli (b), (e) first m is Exercise 2.17. 10.2. meter, second is milli (c), (d) meter Exercise 2.18. (a) (i) 0:42 O, (ii) 4:08 O; (b) (i) 0:33 O, : O Exercise 1.9. Av ffi 1=b; b determines the gain. (ii) 3 23 Exercise 1.10. x ffi 5:064. Exercise 2.19. Exercise 1.11. (a) 0:495, (b) 5, (c) 1, (d) 3:02, (e) 50. 10 Exercise 1.12. (a) i0, (b) 0. Exercise 1.13. (a) 2:015 Â 103, (b) 16:38 Â 106, 1 (c) 759 Â 10 3, (d) 462 Â 10 6, (e) 4:792 Â 103.
0.1 2. Chapter 2
0.01 = Exercise 2.1. I ¼ 2Nqp T where qp is the charge of a 1 proton. SI 2Nqp=T ¼ Cs ¼ A. 0.001 3 4 5 6 7 8 9 Exercise 2.2. h ¼ u2=ð2gÞ;SI½ ¼u2=ð2gÞ m. 10 10 10 10 10 10 10 f pffiffiffiffiffiffiffiffiffiffiffi hipffiffiffiffiffiffiffiffiffiffiffi (Hz) d d Exercise 2.3. q ¼ d f =k;SI d f =k ¼ C. 20 (mm) 100 (mm)
T.H. Glisson, Introduction to Circuit Analysis and Design, 745 DOI 10.1007/978-90-481-9443-8, # Springer ScienceþBusiness Media B.V. 2011 746 Appendix: Answers to Exercises
3. Chapter 3 Exercise 4.2. 2 2 R1ð2R0 þ 4R0R2 þ R2 Þ Req ¼ 2 2 . 2R0 þ 4R0R2 þ R2 þ 2R1ðÞR0 þ R2 Exercise 3.1. vbd ¼ 15 V; vcd ¼ 5V; vda ¼ 20 V; v1 ¼ 5V; vb ¼ 15V. Exercise 4.3. a ¼ 2; b ¼ 19=5; g ¼ 6=5; d ¼ 1. Exercise 3.2. (a) The series connection of elements Exercise 4.4. 8:77 O. 1 and 2 is in parallel with the series connection of R0R1R2 elements 3 and 4. (b) The parallel connection of Exercise 4.5. (a) vx ¼ ðÞi0 i1 R ðÞþR þ R R R elements 1, 2, and 3 is in series with element 4. 1 0 2 0 2 R1R2 : % Exercise 3.3. vab ¼ v1 v2 v3, vac ¼ v1 v3, vbd ¼ (b) vx ffi i1 (c) 1 06 . R1 þ R2 v þ v . 2 3 : : : Exercise 3.4. No. i ¼ i þi so the sources aren’t Exercise 4.6. (a) RT ¼ 0 91R, vT ¼ 0 91Ri0 0 48v0 2 1 3 : : = : independent. (b) RN ¼ 0 91R, iN ¼ i0 0 53ðÞv0 R RnRm ; R23I0 ; Exercise 3.5. Let Rnm ¼ I1 ¼ 4v1 þ 12Ri0 12 Rn þ Rm R1 þ R23 Exercise 4.7. vT ¼ v0 þ , RT ¼ R: R13I0 R12I0 7 7 I2 ¼ ; I3 ¼ . R2 þ R13 R12 þ R3 7v0 þ 4v1 12 Exercise 4.8. iN ¼ i0 þ ; RN ¼ R. R1 12R 7 Exercise 3.6. v1 ¼ v0. R1 þ R2 ðÞR2 R1 v2i1 R2v2 R1R2i1 Exercise 3.7. Exercise 4.9. vT ¼ , RT ¼ . R2i1 v2 R2i1 v2 R3v2 ðÞR2 þR3 v1 R2 R3v1 ðÞR1 þR3 v2 i1 ¼ ; i3 ¼ . R R þR R þR R R R R þR R þR R R2ðÞ3R1 þ 2R2 1 2 1 3 2 3 3 1 2 1 3 2 3 Exercise 4.10. RN ¼ . R1 þ R2 Exercise 3.8. vac ffi 0:015v0; vbc ffi 0:463v0. va va ðÞvd v2 v1 Exercise 3.9.ðÞ a i1 þ þ ¼ 0; R1 R2 v v v v v v v 5. Chapter 5 ðÞd d þ d 2 þ d 2 1 a ¼ 0. R4 R3 R2 v R i v v þ v Exercise 3.10. ðbÞ b 1 1 þ c þ c 2 ¼ 0; Exercise 5.1. Answer given in problem statement. R1 þ R2 R3 R4 vc vc þ v2 vb R1i1 Exercise 5.2. ðcÞ þ þ ¼ 0. R3 R4 R1 þ R2 2 2 Exercise 3.11. Choose node c as reference node. pR ¼ð1=9RÞðÞV0 2RI0 cos ðÞo0 t 0 R R R i i v 1 2 3ðÞþ1 þ 2 1 2 2 vb ¼ . p2R ¼ð2=9RÞðV0 þ RI0Þ cos ðÞo0 t 0 R1R2 þ R1R3 þ R2R3
Exercise 3.12. No. v3 ¼ v1þv2 so the sources aren’t 2 2 pi ¼ ð2=3ÞðV0I0 þ RI0 Þ cos ðÞo0 t 0 independent. R R v R v ðÞ2 þ 3 a 3 b 2 2 Exercise 3.13. i1 ¼ . pv ¼ ð1=3RÞðV0 2RI0V0Þ cos ðÞo0 t R1R2 þ R1R3 þ R2R3 Exercise 3.14. p > 0 and p > 0 so the resistors consume energy. R v R R R R i R 2R B 0 þ ðÞ2 3 2 4 0 ; < i1 ¼ 2 pi 0, so the current source produces energy. If RARB R2 2 V0 2RI0V0 > 0, then Pv < 0 and the voltage source R2v0 þ RAðÞR3 5R4 i0 2 ; produces energy. If V0 2RI0V0 < 0, then pv > 0 and i2 ¼ 2 RARB R2 the voltage source consumes energy. RA ¼ R1 þ R2; RB ¼ R2 þ R3 þ R4: Exercise 5.3. P ¼ 25 mW; V ¼ 5V; w ¼ 50 mJ. Exercise 5.4. P ffi 547 mW: Exercise 3.15. Answer given in problem statement. Exercise 5.5. Answer given in problem statement. Exercise 5.6. 4. Chapter 4 P y p^ ¼ 0 ½ ¼1 þ cosðÞy P cos2 2 0 2 3 1 3v0 Exercise 4.1. ia ¼ þ va þ i0. Exercise 5.7. Answer given in problem statement. 2R0 R1 2R0 Appendix: Answers to Exercises 747
Exercise 5.8. 6. Chapter 6
RS V + I R S – S S Exercise 6.1. Rin ¼ R1kðÞR2 þ R3kRL ¼ R1ðÞR2R3 þ R2RL þ R3RL VS VS = 12.6 V, RS = 0.1 Ω IS == 126 A, RS = 0.1 Ω R1R3 þ R1RL þ R2R3 þ R2RL þ R3RL RS 2 2 Rout ¼ R3kðÞR2 þ R1kRS ¼ VS IS RS P = = ≅ 397 W; I = I = 126 A L max R L max S R ðÞR R þ R R þ R R 4 S 4 3 2 1 2 S 1 S , R1R3 þ R1RS þ R2R1 þ R2RS þ R3RS Exercise 5.9. There are two solutions: (b) and (c) same as (a). 179O 189V 1:06A Exercise 6.2. g ¼ m=R. R ffi ; V ffi ) I ffi . T 0:557O T 10:6V N 19:0A RoRL RiRS Exercise 5.10. 1=2. Exercise 6.3. vab ¼ gis. Ro þ RL Ri þ RS Exercise 5.11. 1 þ cosðÞy1 y2 . R1 Exercise 5.12. Let x ðÞ¼t x ðÞ¼t cosðÞ)o t Exercise 6.4. Rin ¼ . 1 2 1 þ b 2 1 ; x1x2 ¼ cos ðÞ¼o t 2 x1 x2 ¼ cosðÞo t cosðÞ¼o t 0. RoRL RiRS ; Exercise 5.13. 1. Exercise 6.5. vT ¼ gis 2 2 Ro þ RL Ri þ RS I0R I0R Exercise 5.14. ptðÞ¼ þ cosð4pftÞ; p^¼ 1:5kW; RoRL 2 2 RT ¼ . P ¼ 750W. Ro þ RL Exercise 5.15. Exercise 6.6. R2 R2 2; 2; 2 ðÞ7m 1 v ðÞ1 þ 4m v þ ðÞ11b 4m 1 Ri PR1 ¼ ðÞi0 i1 PR2 ¼ ðÞi0 i1 ðÞi0 i1 1 2 1 R1 R2 vout ¼ . m 19 1 2 2 ¼ I0 þI1 2I0I1 cosðÞy 2 Exercise 6.7. m1 ¼ 0. I2R I I R I2R I I R Exercise 6.8. m ¼ gR0. P ¼ 0 þ 0 1 cosðÞy ; P ¼ 1 þ 0 1 cosðÞy : i0 2 2 i1 2 2 Exercise 6.9. Exercise 5.16. ðaÞ 25 mA ðbÞ 5VðcÞ p10ffiffi V ðdÞ p10ffiffi mA. 2 2 Ri Rf ðÞþRL þ Ro RLRo Rin ¼ ; Exercise 5.17. Rf ðÞþRL þ Ro RLRo þ Ri½ Ro þ RLðÞm þ 1 2 2 ÀÁ V0 þ ðÞI0R ; 1 2 ; PR ¼ Pi ¼ V0I0 þ I0R Ro Rf ðÞþRi þ RS RiRS 4R 4 R ¼ ÀÁ : 2 out V V I R Rf þ Ro ðÞþRi þ RS ðÞm þ 1 RiRS P ¼ 0 0 0 ; P þ P þ P ¼ 0: v 4R R i v 7% Exercise 5.18. (a) 4 s, (b) 10 . pffiffiffi Exercise 6.10. Exercise 5.19. Yes. Measure peak and divide by 2. Ro Exercise 5.20. 1:037 kO. + Exercise 5.21. The load voltages are equal and given v R + mv 1 i – 1 V – by V ¼ 0 : L 2 2 2 Ro1Ro2 V0 ðÞR þR0 V0 Exercise 5.22. P ; P (a) Ri ¼ Ri1, Ro ¼ Ro2, m ¼ b2g1, max ¼ V0 ¼ R R 4RR0 2R0 o1 þ i2 R R 2R V2 R 2R 2 V2R o1 þ 0 ; 0 þ 0 ; 0 (b) VCCS: Ri ¼ Ri1, Ro ¼ Ro2, g ¼ b2g1, PR0 ¼ PR ¼ . R þ R 4 2 o1 i2 R þR0 R0 R þR0 4ðÞR þ R0 R R R CCVS: R ¼ R , R ¼ R , r ¼ i1 o1 o2 b g Exercise 5.23. 0.5. ÀÁ i i1 o o2 R R 2 1 2 2 2 o1 þ i2 2RL R0I0 þ V0 Exercise 5.24. P ¼ ¼ 25 mW: Ri1Ro1 L 2 CCCS: R ¼ R , R ¼ R , b ¼ b g . ðÞR0 þ R1 þ 2RL i i1 o o2 2 1 Ro1 þ Ri2 748 Appendix: Answers to Exercises Exercise 6.11. 4RS RiRoRLg APdB ¼ 10 log þ 20 log : RL D RT = 7.5 Ω Exercise 6.31. 114 dB. + V = 15 V I = 2A R = 7.5 Ω – T N N 7. Chapter 7
Exercise 6.12. No. Magnitudes only. Exercise 6.13. 7.5 W. Exercise 7.1. (a) (ii), (b) (iv), (c) (i), (d) (iii). Exercise 6.14. Two. Exercise 7.2. (a) (iii), (b) (iv), (c) (i), (d) (ii). Exercise 6.15. RL=RS 19: R4 R2 Exercise 7.3. vL ¼ 1 þ vS. Exercise 6.16. RL=RS 0:053: R3 R1 Exercise 6.17. 8=9 ffi 0:889: Exercise 7.4. vL ¼ v1 þ v2. Exercise 6.18. 0:63 RL=RS 1:58: Exercise 7.5. Rin¼ Exercise 6.19. D ¼ ðÞRi þ RS ðÞRo þ RL ðÞm0þ1 RLRiR1þðÞR1þRi ½ þR2ðÞþRLþRo RoRL RiR1Ro: mRiRL mRiRSRL R ðÞþR þR R ðÞþR þR R R Hv ¼ , Hr ¼ ; 2 o L L o 1 1 o D D Av Exercise 7.6. Rout ffi Ro: mRi mRiRS m0 þ Av Hg ¼ ; Hi ¼ : D D Exercise 7.7. Exact: Rin ¼ 974 GO; Rout ¼ 808 mO O; mO Exercise 6.20. D ¼ ðÞRS þ Ri ðÞRo þ RL , Approx: Rin ¼ 990 G Rout ¼ 808 . rR rR Exercise 7.8. Answer given in problem statement. A ¼ L , A ¼ S , ; v D i D Exercise 7.9. Inverting amp: Exact Av ¼ 99 Rin ¼ 10 kO; Rout ¼ 10:01 mO. r rRSRL Ag ¼ , Ar ¼ . Approx: A ffi 99:01; R ffi 10 kO; R ffi 9:9mO. D D v in out Non-inverting amp: Av ¼ 101; Rin ¼ 97:1GO; Rout ¼ Exercise 6.21. D ¼ ðÞRS þ Ri1 ðÞRo1 þ Ri2 ðÞRo2 þ RL 10:1mO rbR R rbR R Approx: Av ffi 101; Rin ffi 99 GO; Rout ffi 10:1mO: A ¼ o2 L ; A ¼ S o2 ; v D i D Exercise 7.10. Answer given in problem statement. rbR rbR R R Exercise 7.11. PA max ¼ 78:1 mW (virtually any op A ¼ o2 ; A ¼ S o2 L : g D r D amp will satisfy this requirement). Exercise 7.12. R0 ¼ R kðÞffiR þ R 998 O; = = L L 1 2 Exercise 6.22. Av ¼ AiRL RS, Ag ¼ Ai RS, Ar ¼ RLAi. V2 = ; = ; CC Exercise 6.23. (a) a ¼ 1 RL (b) a ¼ 1 RS PA max ¼ 0 ffi 157 mW. 4RL (c) a ¼ 1=RS; (d) a ¼ 1: Exercise 6.24. Results follow immediately from rela- tions among the gains (see Example 7.12). Exercise 6.25. Usually no, because increasing the out- 8. Chapter 8 put resistance of the source increases the power wasted (decreases the power transferred to the load). Exercise 8.1. Exercise 6.26. No, they are consistent because A ¼ i v RSAv=RL. 10 Exercise 6.27. AP ¼ 6:31 Â 10 : Exercise 6.28. PA ¼ 0:5mW: Exercise 6.29. Follows immediately from Exercise 6.24. Exercise 6.30. D ¼ ðÞRo þ RL ðÞRS þ Ri RiRoRLg RSRiRog AvdB ¼ 20 log ; AidB ¼ 20 log ; 0 t D D 0 Appendix: Answers to Exercises 749
Exercise 8.2.
v (t) dv 5V 0 ≤ t < 4ms: C = (20 pF) = 25 nV dt 4ms 50 nV dv 4ms ≤ t < 5ms: C = 0 25 nV dt dv –10 V t (ms) 5ms ≤ t < 7ms: C = (20 pF) = –100 nV dt 2ms dv 5V 10ms ≤ t < 12ms: C = (20 pF) = 50 nV dt 2ms
–100 nV
Exercise 8.3. itðÞ¼2pfVCcosðÞ 2pft , maxjj¼itðÞ 2pfVC¼ 31:42A: Exercise 8.4.
v(t)(V)
6 V 1 ∫ t i t ′ dt ′ = 10 mA × t = × 6 × t ≤ t < μ 0 ( ) 2 10 ; 0 3 s 4 C 5 nF s etc...
12 t (µs) 3 4 6 8
–2
Exercise 8.5. lim vCðÞ¼t RI0: Exercise 8.10. vC; i2. t!1 Exercise 8.6. The model is unrealistic if i0 has a Exercise 8.11. 4C. dc component, because the charge on the capacitor Exercise 8.12. No, because the models would still allow would grow without bound (mathematically). Also, if capacitor voltages to change instantaneously. the voltage vC is the quantity of interest, the resistor is Exercise 8.13. irrelevant, and one must wonder why it is in the X5R ) 55Cto85C; Æ 15%; model. We cannot find the voltage vCðÞt1 at any Y5V ) 30Cto85C; þ 22%; 82%; time t1 unless we know (or are given) vtðÞ0 for some Z5U ) 10Cto85C; þ 22%; 56%: time t t andanexpressionforthecurrenti ðÞt for 0 1 0 Exercise 8.14. Any combination ______, where the t t t . 0 1 first character is X, or Y, the second is 5 or 7, and the O: Exercise 8.7. R ffi 101 third is A,B,C,D,E,F,P,R,S,T, or U. : : Exercise 8.8. C ffi 39 1pF Exercise 8.15. m ; m : Exercise 8.9. ffi 50 sat1 ffi 250 sat2 0.3 5
4.5
4
3.5 0.2
3 w (fJ ) v (V) 2.5
2 0.1 1.5
1
0.5
0 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 0 0123456789101112 t (µs) t (ms) 750 Appendix: Answers to Exercises pffiffiffi Exercise 8.16. t¼RC; expðÞ¼ t=t 1= 2 )tffi866ns. Exercise 9.4. vtðÞ¼2 p fI1 cosð2 p ftþ p=4Þ: Exercise 8.17. v ðÞ¼t 14:14 þ31:6cosðÞ 2pf t V; f ¼ C 0 0 Exercise 9.5. À ÀÁ 20kHz: t=t itðÞ¼0; t < 0; itðÞ¼ V0t=LÞ 1 e ; t 0: : : Exercise 8.18. (a) ffi 10 9 MHz, (b) ffi 11 V Exercise 9.6. Exercise 8.19. i (t) (µA) 0; t 0 1.5 vout ¼ V0 sinðÞ 2 p ft =ð2 p fRCÞ; t > 0 1.0
Exercise 8.20. (a) VCC 25 V, (b) ffi 5ns: Exercise 8.21. C ffi 563 mF. 12 0 t (µs) V 3684 Exercise 8.22. V ¼ IR; 99I ¼ 99 ¼ CVo0 ) C ¼ 99 R ffi 158 mF: –0.5 o0R Exercise 8.23. f <18:4Hz: Exercise 9.7. Answer implied by problem statement. R2V0 t Exercise 8.24. Exercise 9.8. vL ¼ exp . Other ans- ðÞR1 þ R2 t wers implied by problem statement. Td : Exercise 9.9. t ¼ D = lnðÞ 1 2 V Vmax t L iLðÞ¼t I0; t 0; iLðÞ¼t I0 exp ; t > 0; t ¼ t R LI t v ðÞ¼t 0; t 0; v ðÞ¼ t 0 exp ; t > 0: L L t t 9. Chapter 9 Exercise 9.10. L. 1 Exercise 9.11. wtðÞ¼ LI2½ 1 þ cosðÞ 4pft : 4 Exercise 9.1. H. Exercise 9.12. 0 V; V =R: Exercise 9.2. 97:18 mm; 0:724 O: 0 Exercise 9.3. Exercise 9.13. 0 V; I0: Exercise 9.14. f ¼ 0:01R=ðÞ2pL : 6 Exercise 9.15. Vac rms ffi 1:2mV; g ffi 2:65 Â 10 : di 5mA Exercise 9.16. L ffi 31:8mH: L = (10mH)=12.5mV; 0 £ t <4ms, etc... dt 4ms Exercise 9.17. Answer implied by problem v(t) statement. Exercise 9.18. (a) Answer implied by problem statement. 25 mV (b) No. i1 could have dc component. 12.5 mV Exercise 9.19. (a) Dots at tops of coils. (b) Reverse 57 t (ms) one winding. pffiffiffiffiffiffiffiffiffiffiffiffiffi = : 0 4 10 12 Exercise 9.20. n ¼ RpL ffiffiffiffiffiffiffiffiffiffiffiffiffiRS Exercise 9.21. n1n2 ¼ RL=RS: nvðÞ t : Exercise 9.22. itðÞ¼ 2 2R þ n RL –50 mV Appendix: Answers to Exercises 751
10. Chapter 10 Exercise 11.7. Answer implied by problem statement. Exercise 11.8. d !1: Exercise 11.9. Answer implied by problem statement. Exercise 10.1. Answers given in problem statement. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi Exercise 11.10. Exercise 10.2. z real ) Re2ðÞþz Im2ðÞz ¼ z2 ¼jjz . Remaining exercises: answers given in problem a 6 statement.
4
vC (V)
11. Chapter 11 2
Exercise 11.1. itðÞ¼I0 þ ðÞI1 I0 utðÞþ t0 0 0 20 40 60 80 100 120 140 160 I I ut t : ðÞ2 1 ðÞ 1 b t (µs) V0 ; ; : 5 Exercise 11.2. IL ¼ VC1 ¼ R2I VC2 ¼ 0 R1 þ R2 Exercise 11.3. vLðÞ¼t vLðÞ1 ½ 1 expðÞ t=t utðÞ; t ffi 148:52 ms; vLðÞffi1 334:75 V: vC (V) 0
400 –5 0 20 40 60 80 100 120 140 160 c t (µs) 300 10
vL (V) 200 0
vC (V)
–10 100
–20 0 0 20 40 60 80 100 120 140 160 –200 0 200 400 600 800 t (µs) t (µs) Exercise 11.11.
2 d vC L dvC 3 2 3 2 Exercise 11.4. LC þ þ vC ¼ 0; t > 0; d vo d vo dvo d vS d vS dvS dt2 R dt 4 þ14 þ8 þvo ¼4 þ4 þ4 þvS SI½ ¼LC s2; SI½ ¼L=R s: dt3 dt2 dt dt3 dt2 dt 0 1 Exercise 11.5. (a) R ffi 79:1kO; R < R ) 1 0 1 0 0 B C 2:823 = B 8 C @ A overdamped (b) R ¼ 10R0 ) vCðÞ¼t 2jjY exp ðÞ t t v:¼ st :¼ PolyrootsðÞ v st ¼ 0:5 : @14A cosðÞo tþy utðÞ; 2jjY ffi503mV;tffi1:58ms; 0:177 0 4 6 1 o0 ffi 6:29 Â 10 s ; y ffi 1:57; R ¼ R0 ) vCðÞ¼t Yt expðÞ t=t utðÞ; Y ffi 6:33 Â 107 Vs 1; t ffi 158ns Characteristic roots real and negative, circuit is over- damped. The circuit cannot be underdamped because R ¼ R0=10 ) vCðÞ¼t Y½expðÞ t=t1 expðÞ t=t2 utðÞ; Y ffi 5:03V; t ffi 3:16ms; t ffi 7:93ns: it consists of only resistors and capacitors. 1 2 t ; : ; ; Largest time constant is t ffi ffi5:65t¼ 56:5s: Exercise 11.6. n cycles ffi 1 1 6 4 8; Yes. 3 0:177 752 Appendix: Answers to Exercises
12. Chapter 12 Exercise 12.12. I~ffi 6:37ff p=2mA ) itðÞffi6:37 cosðÞo t p=2 mA. Exercise 12.1. 2p=3. Im Exercise 12.2. Lags. ~ ; m ; V Exercise 12.3. v2 ¼ V2 cosðÞ 2pftþ y V2 ¼ 500 V Re f ¼ 1 kHz; y ffi 1:26: Exercise 12.4. 50ff ðÞp=4 mA: ~ Exercise 12.5. V ¼ V0ff ðÞf p=2 : ~ Exercise 12.6. I¼50ffp=3mAffiðÞ25þj43:3 mA; leads: I~ Exercise 12.7. itðÞ¼I0 cosðÞ 2pftþ y ; I0 ¼ 50 mA; f ¼ 4 MHz; y ¼ p=3: O: Exercise 12.8. Exercise 12.13. 178pffiffiffiffiffi k : : I~ Exercise 12.14. 10 ffi 3 16 Im Exercise 12.15. vðtÞffi146 cosðÞo t 0:350 mV: ~ ~ ~ Exercise 12.16. I ¼ I1 þ I2 ffi 14:1ff 0:065 mA ) itðÞffi14:1 cosðÞo t 0:065 mA; f ¼ 100 kHz: 2 ðÞoL R1 R3 Exercise 12.17. Z¼R2 þ 2 þ 2 þ p /3 V~ R2 oL 1 oCR 2 1 þðÞ þðÞ3 Re "# –p /4 oLR2 oCR2 j 1 3 : 2 2 2 R1 þðÞoL 1þðÞoCR3 ~ 2 ~ V 1 o LCþjoCRðÞ1 þR2 Exercise 12.18. I¼ ¼ V0; V~ Z ðÞR1 þjoL ðÞ1þjoCR2 1 itðÞ¼I0 cosðÞotþy ; I0 ¼1:14mA; y¼ 0:44: Exercise 12.19. f ¼ 5 kHz; R ffi 469 O; L ffi 3:85 mH; Exercise 12.9. Answer implied by problem statement. f ¼ 1 kHz; R ffi 540 O; C ffi 52:7 mF: Exercise 12.10. Answer implied by problem statement. Exercise 12.20. f ¼ 5 kHz; R ffi 500 O; L ffi 61:7 mH; ~ : = Exercise 12.11. I ffi 20 7ffp 2mA f ¼ 1 kHz; R ffi 540 O; C ffi 1:65 nF: : : ) itðÞffi 20 7 sinðÞ 2pf0 t mA Exercise 12.21.
Im (Z) (Ω) 540 Re (Z )(Ω) 121
–3.02 Re (Z )(Ω) Im (Z)(Ω) 469 f = 5 kHz f = 1 kHz
1.20 Re (Y ) (mS) Im (Y) (mS) 0.010
–0.516 Re (Y ) (mS) Im (Y ) (mS) 1.85 f = 5 kHz f = 1 kHz Appendix: Answers to Exercises 753
Exercise 12.22. and RL ¼ RT if RT 6¼ 0. If RT ¼ 0 (not realistic), then ~ VT ffi ðÞ2:47ff 0:251 V; ZT ffi ðÞ107ff 1:34 O: RL should be the value that draws the maximum cur- Exercise 12.23. Resonant frequency is rent from the (current-limited) source. R þ R Exercise 13.9. (a) Answer implied by problem statement. o ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 2 if R2C > L. Otherwise circuit is 2 2 1 (b) No, (c) Yes. LR1C L not resonant. Exercise 13.10. (a) f ¼ 0ðdcÞ (b). Exercise 12.24. Two resonant frequencies: f1 ffi 1.25 67:1kHz; f2 ffi 13:3kHz:
1 13. Chapter 13
0.75 Exercise 13.1. Load: S ffi ðÞ112þj298 VA; P ffi 112W Source impedance: S ffi ðÞ90:1þj68:0 VA; P ffi 90:1W 0.5 Voltage source: S ffi ðÞ202 þ j365 VA; P ffi 202 W:
Exercise 13.2. Source: ffi 790 VAR, Capacitor: 0.25 ffi 802 VAR, Inductor: ffi 12:2 VAR: Exercise 13.3. pf ¼ 1. 0 Exercise 13.4. See figure below. Each side of the 105 106 107 108 109 2 f power triangle equals Vrms times the corresponding (Hz) h f side of the admittance triangle. Pn ( f ) ( ) Exercise 13.5. (a) ffi 17:5 kVAR (b) pf ffi 0:87. Exercise 13.11. 18 dB.
2 Q = V 2 Y 14. Chapter 14 |S | =Vrms |Y | rms Im ( ) 2 = BYVrms
q Exercise 14.1. Answer implied by problem statement. Exercise 14.2. No. P = V 2 Re (Y ) = G V 2 rms Y rms Exercise 14.3. Both are zero. (a) S = P + jQ
|Y| 15. Chapter 15
Im (Y ) = BY : : q Exercise 15.1. vL ffi 8 40 cosðÞo1 t þ 0 82 þ12:23 cosðÞþo2 t 0:06 9:00 cosðÞo3 t 0:75 : Re (Y ) = GY Exercise 15.2. (b) Y = GY + jBY
Hv Hi Hz Hy 1 1 Exercise 13.6. Answer implied by problem statement Hv 1 ZS ZL ZS ZL 1 1 Exercise 13.7. Answer implied by problem statement Hi ZL ZS 1 ZL ZS Hz ZS ZL 1 ZLZS Exercise 13.8. The equivalent series reactances XL and 1 1 1 Hy ZL ZS ðÞZLZS 1 XT must be non-negative (capacitor doesn’t pass dc) 754 Appendix: Answers to Exercises
Exercise 15.3. Current transfer: Zin ! 0; Zout !1; Exercise 16.7. Transimpedance: Z ! 0; Z ! 0; Transadmittance: in out f ðÞkHz 4 8 12 16 20 24 28 32 Zin !1; Zout !1. Ak ðÞmA 0.00 45.02 31.83 15.01 0.00 9.00 10.61 6.43 ZS K Exercise 15.4. HiðÞ¼jo1 . yk 0.00 0.00 0.00 0.00 0.00 3.14 3.14 3.14 ZLðÞjo1 1 þ jo1=o0 Exercise 15.5. 10 W m 1. Exercise 16.8. Answer implied in problem statement. Exercise 15.6. A ffi 33 dB; A ffi 50 dB. v i Exercise 16.9. a ¼ x ; b ¼ 0: Exercise 15.7. jjffiY 17:8mS: 0 dc 0 Exercise 16.10. Exercise 15.8. Because 1 Æ f =f0 is not a linear factor (no j). K0123
ak ðÞV 2.0 1.0 0.4 0.2 bk ðÞV 0.0 1.0 0.8 0.6 16. Chapter 16 V0 ; ; ; ; ; Exercise 16.11. Ak ¼ 2 yk ¼ kp k ¼ 1 2 ÁÁÁ V k X ¼ x ; X 0 ðÞ 1 k; k ¼Æ1; Æ2; ÁÁÁ; a ¼ x ; Exercise 16.1. Average value: xdc ¼ A0ffy0 ¼ 500 mV 0 dc k 2k2 0 dc 1 V0 k Third harmonic: f ¼ 3f ¼ 6kHz; T ¼ ffi 167ms; b0 ¼ 0; ak ¼ ðÞ 1 ; bk ¼ 0; k ¼ 1; 2; ÁÁÁ: 0 3 k2 500 3p p f0 A3 ¼ pffiffiffiffiffi mV; y3 ¼ ¼ : Exercise 16.12. 10 6 2 A5 500 Fifth harmonic: A5 rms ¼ pffiffiffi ¼ pffiffiffipffiffiffiffiffi mV: 2 2 26 k 0123 p 500 Ak ðÞmV 1.00 1.41 0.89 0.63 Fundamental: f0 ¼ 2 kHz; y1 ¼ ; A1 rms ¼ pffiffiffipffiffiffi 6 yk 0.00 0.52 1.05 1.57 mV 250 mV: 2 2 ¼ ak ðÞmV 1.00 1.23 0.45 0.00 : Exercise 16.2. 5f0 ¼ 1 5 kHz ) f0 ¼ 300 Hz ) T ¼ bk ðÞmV 0.00 0.71 0.78 0.63 1 ffi 3:33 ms: f kp 0 Exercise 16.13. X ¼ V sa : k ¼ 0; Æ1; Æ2; ÁÁÁ: Exercise 16.3. Only the third, whose frequencies are k 0 2 harmonics of a fundamental. T t < T t : ; m ; Exercise 16.14. t þ . Exercise 16.4. P1 ¼ 3 125 mW P2 ¼ 500 W P3 ¼ 2 2 2 2 m ; : m ; : m : 125 W P4 ¼ 43 25 W P5 ¼ 18 49 W x0t 2 kpt x0t 2 kt Exercise 16.15. Xk ¼ sa ¼ sinc ; P ¼ 3:812 mW; ¼ 0:994: 2T 2T 2T 2T 2 kp 2 k Xk ¼ x0sa ¼ x0sinc : Exercise 16.5. X0 ¼ 10 mV, X1 ¼ 25 mV, X3 ¼ 2 2 : : : : ðÞ12 5ff 0 785 mV, X5 ¼ ðÞ6 25ff 1 57 mV, all Exercise 16.16. others ¼ 0. Exercise 16.6. Average value ¼ dc component ¼ k 123
X0 ¼ 1V Ak ðÞV 6.37 3.18 2.12 Period T ¼ 2 ms Fundamental frequency yk 1.57 1.57 1.57 1 f ¼ ¼ 500 Hz 0 T Exercise 16.17. Third harmonic: A3 ¼ 2jjX3 ffi 632 mV; R2ðÞ1 þ joR1C 2 kp 1 1 (a) Yk ¼ V0 sa ,(b). y ¼ ffi 1:25; ffi 667 ms R1ðÞ1 þ joR2C 2 3 1 þ j3 3f 0ffiffiffi A5 p Fifth harmonic: pffiffiffi ¼ 2jjX5 ¼ 277 mV k 01 23 45 67 2 Ak ðÞmV 0.00 405.29 0.00 45.03 0.00 16.21 0.00 8.27 Fundamental: f ¼ 500 Hz; y ¼ X ffi 0:785; pffiffiffi pffiffiffi 0 1 1 Bk ðÞV 0.00 1.46 0.00 0.83 0.00 0.02 0.00 0.01 A1= 2 ¼ 2jjX1 ¼ 1V. Appendix: Answers to Exercises 755
Exercise 16.18. 17. Chapter 17 5
4.167 Exercise 17.1. 120 3.333
Ak ()mV 2.5 100 1.667
0.833 Av (dB) 80
0 0 246 8 10 12 14 16 f 60 10Hz 0 40 1 10 100 1.103 – 0.1 f (Hz)
– 0.2 θk π Exercise 17.2. Answer implied by problem statement. – 0.3 pffiffiffi Exercise 17.3. (a) AvðÞ¼0 100; AvðÞ¼W 100 2; O; : O – 0.4 (b) ZinðÞffi0 10 k jjffiZinðÞj2pW 14 14 k ; ZoutðÞ0 ffi 0:038 O; jjZoutðÞj2pW ffi 39:78 O: – 0.5 0 246 8 10 12 14 16 f 10Hz Exercise 16.19.
3
2
Xk ()mV
1
0 –100 0 100 f ()Hz
0.6
0.4
0.2 θ k 0 π –0.2
–0.4
–0.6 –100 0 100 f ()Hz 756 Appendix: Answers to Exercises
Exercise 17.4.
Inverting Amplifier 50 20
40 0
30 –20 Zin Z ()dB o ()dB R R i 20 o –40
10 –60
0 –80 0.1 10 1´103 1´105 1´107 0.1 10 1´103 1´105 1´107 f ()Hz f ()Hz Non-Inverting Amplifier 80 20
60 0
–20 Z 40 Z in ()dB o ()dB R R i 20 o –40
0 –60
–20 –80 0.1 10 1´103 1´105 1´107 0.1 10 1´103 1´105 1´107 f ()Hz f ()Hz Voltage Follower 150 50
100 0 Z in ()dB Z R o ()dB i R 50 o –50
0 –100
–50 –150 0.1 10 1´103 1´105 1´107 0.1 10 1´103 1´105 1´107 f ()Hz f ()Hz
: : Exercise 17.5. C ffi 3 98 pF v(t – t ) (V) : 0 Exercise 17.6. Both ratios ffi 1 5
18. Chapter 18 0 75 125 t (μs) (a) 2 2 Exercise 18.1. FsðÞ¼ 2a=ðÞs a ; jsj Exercise 18.7. Answer implied by problem statement. 25 75 t (μs) Exercise 18.8. (b) Appendix: Answers to Exercises 757 Exercise 18.21. I ¼ b I ¼ b b I ¼ b b I No. 1 expðÞ st ; L 2 2 1 2 1 1 2 S vtðÞ¼V0½ )utðÞ utðÞ t VsðÞ¼V0 I s L 1 expðÞ st : I R I I R LfgvtðÞ t0 ¼ V0 expðÞ st0 S S 1 2 L s b I b I 1 1 2 2