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Appendix: Answers to Exercises Appendix: Answers to Exercises pffiffiffiffiffi 1 1 1. Chapter 1 Exercise 2.4. FðÞ¼x x kf þ ; 1 0 x x À x pffiffiffiffiffi 1 1 1 1 0 SI x kf À ¼ V. 0 x x x Exercise 1.1. Physical law. 1 1 À 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 À xavab kq À xavab Exercise 2.6. i2 ¼10 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)Âà 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 SI½¼i 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) 10À4 AsÀ1, (b) 104 VsÀ1, (c) 105 VA, Exercise 2.13. Various answers (d) 1010msÀ2. m Exercise 2.14. Because 298 K ¼ 25 C, 298 K is the À1 5 V 5nV m Exercise 1.7. (a) 5mVs ¼ ms ¼ ms , (b) 25kA s¼ reference temperature. À1 À1 À1 25Ams¼25MAns, (c) 100 mJ ms ¼ 100mJms ¼ Exercise 2.15. a10 ¼ 0:0042 K . 100 JsÀ1. 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  103, (b) 16:38  106, 1 (c) 759  10À3, (d) 462  10À6, (e) 4:792  103. 0.1 2. Chapter 2 0.01 = Exercise 2.1.ÂÃI ¼ 2Nqp T where qp is the charge of a À1 proton. SI 2Nqp=T ¼ Cs ¼ A. 0.001 3 4 5 6 7 8 9 Exercise 2.2. h ¼ u2=ð2gÞ;SI½¼u2=ð2gÞ m. 10 10 10 10 10 10 10 f pffiffiffiffiffiffiffiffiffiffiffi hipffiffiffiffiffiffiffiffiffiffiffi (Hz) d d Exercise 2.3. q ¼ d Àf =k;SI d Àf =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 ¼20 V; v1 ¼5V; vb ¼ 15V. Exercise 4.3. a ¼2; b ¼ 19=5; g ¼ 6=5; d ¼1. 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 À 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 À v2 À v3, vac ¼ v1 À 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 91Ri0À 0 48v0 2 1 3 : : = : independent. (b) RN ¼ 0 91R, iN ¼ i0 À 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 À R1 v2i1 R2v2 À R1R2i1 Exercise 3.7. Exercise 4.9. vT ¼ , RT ¼ . R2i1 À v2 R2i1 À v2 R3v2 ÀðÞR2 þR3 v1 R2 R3v1 ÀðÞR1 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 ffi0:015v0; vbc ffi 0:463v0. va va À ðÞvd À v2 À v1 Exercise 3.9.ðÞÀ a i1 þ þ ¼ 0; R1 R2 v v À v v À v À v À v 5. Chapter 5 ðÞd d þ d 2 þ d 2 1 a ¼ 0. R4 R3 R2 v À 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 À 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 À 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 ¼ð2=3ÞðV0I0 þ RI0 Þ cos ðÞo0 t 0 independent. R R v R v ðÞ2 þ 3 a À 3 b 2 2 Exercise 3.13. i1 ¼ . pv ¼ð1=3RÞðV0 À 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 À 2 4 0 ; < i1 ¼ 2 pi 0, so the current source produces energy. If RARB À R2 2 V0 À2RI0V0 > 0, then Pv < 0 and the voltage source R2v0 þ RAðÞR3 À 5R4 i0 2 ; produces energy. If V0 À2RI0V0 < 0, then pv > 0 and i2 ¼ 2 RARB À 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 ½¼1 þ cosðÞy P cos2 2 0 2 3 1 3v0 Exercise 4.1. ia ¼ þ va þ À 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 À 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 À 1 v À ðÞ1 þ 4m v þ ðÞ11b À 4m À 1 Ri PR1 ¼ ðÞi0 Ài1 PR2 ¼ ðÞi0 Ài1 ðÞi0 Ài1 1 2 1 R1 R2 vout ¼ . Âà m À 19 1 2 2 ¼ I0 þI1 À2I0I1 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. Âà 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 Ri½Ro þ RLðÞm þ 1 2 2 ÀÁ Âà 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 À 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 À7% Exercise 5.18.
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