Tài liệu Bài giải phần giải mạch P6 pptx

Chapter 6, Solution 1. ( =+−== −− t3t3 e6e25 dt dv Ci ) 10(1 - 3t)e -3t A p = vi = 10(1-3t)e -3t ⋅ 2t e -3t = 20t(1 - 3t)e -6t W Chapter 6, Solution 2. 22 11 )120)(40( 2 1 Cv 2 1 w == w 2 = 22 1 )80)(40( 2 1 2 =Cv 1 ( ) =−=−=∆ 22 21 8012020www 160 kW Chapter 6, Solution 3. i = C = − = − 5 160280 10x40 d t dv 3 480 mA Chapter 6, Solution 4. )0(vidt C 1 v t o += ∫ ∫ +1tdt4sin6 2 1 = = 1 - 0.75 cos 4t Chapter 6, Solution 5. v = ∫ + t o )0(vidt C 1 For 0 < t < 1, i = 4t, ∫ − = t o 6 t4 10x20 1 v dt + 0 = 100t 2 kV v(1) = 100 kV For 1 < t < 2, i = 8 - 4t, ∫ +−= − t 1 6 )1(vdt)t48( 10x20 1 v = 100 (4t - t 2 - 3) + 100 kV Thus v (t) =     <<−− << 2t1,kV)2tt4(100 1t0,kVt100 2 2 Chapter 6, Solution 6. 6 10x30 dt dv Ci − == x slope of the waveform. For example, for 0 < t < 2, 3 10x2 10 d t dv − = i = mA150 10x2 10 x10x30 d t dv 3 6 == − − C Thus the current i is sketched below. t (msec) 150 12 10 2 8 6 4 -150 i(t) (mA) Chapter 6, Solution 7. ∫∫ +=+= − − t o 3 3 o 10dt10tx4 10x50 1 )t(vidt C 1 v = =+10 50 t2 2 0.04k 2 + 10 V Chapter 6, Solution 8. (a) tt BCeACe dt dv C 600100 600100 −− −−== i (1) BABCACi 656001002)0( − − =  →  −−== (2) BAvv +=→= −+ 50)0()0( (3) Solving (2) and (3) leads to A=61, B=-11 (b) J 52500104 2 1 )0( 2 1 32 === − xxxCvEnergy (c ) From (1), A 4.264.241041160010461100 60010060031003 tttt eeexxxexxxi −−−−−− −−=−−= Chapter 6, Solution 9. v(t) = () ( ) ∫ −− +=+− t o tt Vet120dte16 21 1 v(2) = 12(2 + e -2 ) = 25.62 V p = iv = 12 (t + e -t ) 6 (1-e -t ) = 72(t-e -2t ) p(2) = 72(2-e -4 ) = 142.68 W Chapter 6, Solution 10 dt dv x dt dv Ci 3 102 − ==      << << << = s4t316t,-64 s 3t116, s10,16 µ µ µ tt v      << << << = s4t3,16x10- s 3t10, s10,1016 6 6 µ µ µ tx dt dv      << << << = s4t3kA, 32- s 3t10, s10,kA 32 )( µ µ µ t ti Chapter 6, Solution 11. v = ∫ + t o )0(vidt C 1 For 0 < t < 1, ∫ == − − t o 3 6 t10dt10x40 10x4 1 v kV v(1) = 10 kV For 1 < t < 2, kV10)1(vvdt C 1 v t 1 =+= ∫ For 2 < t < 3, ∫ +−= − − t 2 3 6 )2(vdt)10x40( 10x4 1 v = -10t + 30kV Thus v(t) =      <<+− << <<⋅ 3t2,kV30t10 2t1,kV10 1t0,kVt10 Chapter 6, Solution 12. π−π== − 4sin)(4(60x10x3 dt dv Ci 3 t) = - 0.7e π sin 4πt A P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W W = ∫∫ t dt ππ−= t o 8 1 o 8sin6.21pdt = π π π 8 8 6. cos 21 8/1 o = -5.4J Chapter 6, Solution 13. Under dc conditions, the circuit becomes that shown below: i 2 50 Ω 20 Ω + − 60V + v 1 − i 1 30 Ω 10 Ω + v 2 − i 2 = 0, i 1 = 60/(30+10+20) = 1A v 1 = 30i 2 = 30V, v 2 = 60-20i 1 = 40V Thus, v 1 = 30V, v 2 = 40V Chapter 6, Solution 14. (a) C eq = 4C = 120 mF (b) 30 4 C 4 C 1 eq == C eq = 7.5 mF Chapter 6, Solution 15. In parallel, as in Fig. (a), v 1 = v 2 = 100 C 2 + v 2 − C 1 + − 100V + v 2 − C 2 + − v 1 C 1 + v 1 − + − 100V (b) (a) w 20 = == − 262 100x10x20x 2 1 Cv 2 1 0.1J w 30 = = − 26 100x10x30x 2 1 0.15J (b) When they are connected in series as in Fig. (b): ,60100x 50 30 V CC C v 21 2 1 == + = v 2 = 40 w 20 = = − 26 60x10x30x 2 1 36 mJ w 30 = = − 26 4010x30 2 xx 1 24 mJ Chapter 6, Solution 16 F 2030 80 80 14 µ =→= + += C C Cx C eq Chapter 6, Solution 17. (a) 4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. C eq = 3F (b) C eq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6F 1 3 1 6 1 2 1 C 1 eq =++= C eq = 1F Chapter 6, Solution 18. For the capacitors in parallel = 15 + 5 + 40 = 60 µF 1 eq C Hence 10 1 60 1 30 1 20 1 C 1 eq =++= C eq = 10 µF Chapter 6, Solution 19. We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60 µ F. The 60 - µ F capacitor in series with another 60- µ F capacitor gives 30 µ F. 30 + 50 = 80 µ F, 80 + 40 = 120 µ F The circuit is reduced to that shown below. 12 120 12 80 120- µ F capacitor in series with 80 µ F gives (80x120)/200 = 48 48 + 12 = 60 60- µ F capacitor in series with 12 µ F gives (60x12)/72 = 10 µ F Chapter 6, Solution 20. 3 in series with 6 = 6x 3 /(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below: 20 1 6 2 8 6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus C eq = 4 mF Chapter 6, Solution 21. 4µF in series with 12µF = (4x12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF Hence C eq = 2.5µF Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below: a b 40 µ F 60 µF 30 µ F 20 µ F Combining the capacitors in series gives C , where 1 eq 10 1 30 1 20 1 60 1 C 1 1 eq =++= C = 10µF 1 eq Thus C eq = 10 + 40 = 50 µF Chapter 6, Solution 23. (a) 3µF is in series with 6µF 3x6/(9) = 2µF v 4µF = 1/2 x 120 = 60V v 2µF = 60V v 6µF = =( 3 + )60 36 20V v 3µF = 60 - 20 = 40V (b) Hence w = 1/2 Cv 2 w 4µF = 1/2 x 4 x 10 -6 x 3600 = 7.2mJ w 2µF = 1/2 x 2 x 10 -6 x 3600 = 3.6mJ w 6µF = 1/2 x 6 x 10 -6 x 400 = 1.2mJ w 3µF = 1/2 x 3 x 10 -6 x 1600 = 2.4mJ Chapter 6, Solution 24. 20µF is series with 80µF = 20x80/(100) = 16µF 14µF is parallel with 16µF = 30µF (a) v 30µF = 90V v 60µF = 30V v 14µF = 60V v 20µF = = + 60x 8020 80 48V v 80µF = 60 - 48 = 12V (b) Since w = 2 Cv 2 1 w 30µF = 1/2 x 30 x 10 -6 x 8100 = 121.5mJ w 60µF = 1/2 x 60 x 10 -6 x 900 = 27mJ w 14µF = 1/2 x 14 x 10 -6 x 3600 = 25.2mJ w 20µF = 1/2 x 20 x 10 -6 x (48) 2 = 23.04mJ w 80µF = 1/2 x 80 x 10 -6 x 144 = 5.76mJ Chapter 6, Solution 25. (a) For the capacitors in series, Q 1 = Q 2 C 1 v 1 = C 2 v 2 1 2 2 1 C C v v = v s = v 1 + v 2 = 2 1 21 22 1 2 v C CC vv C C + =+ s 21 1 2 v CC C + =v Similarly, s 21 2 1 v CC C v + = (b) For capacitors in parallel v 1 = v 2 = 2 2 1 1 C Q C Q = Q s = Q 1 + Q 2 = 2 2 21 22 2 1 Q C CC QQ C C + =+ or Q 2 = 21 2 CC C + s 21 1 1 Q CC C Q + = i = dt dQ s 21 1 1 i CC C + =i , s 21 2 2 i CC C + =i Chapter 6, Solution 26. (a) C eq = C 1 + C 2 + C 3 = 35µF (b) Q 1 = C 1 v = 5 x 150µC = 0.75mC Q 2 = C 2 v = 10 x 150µC = 1.5mC Q 3 = C 3 v = 20 x 150 = 3mC (c) w = J150x35x 2 1 2 22 eq µ= vC 1 = 393.8mJ