Tài liệu Bài giải mạch P11 ppt

Chapter 11, Solution 1. )t50cos(160)t(v = )9018030t50cos(2)30t50sin(20-)t(i ° − ° + ° − =°−= )60t50cos(20)t(i °+= )60t50cos()t50cos()20)(160()t(i)t(v)t(p ° + == [ ] W)60cos()60t100cos(1600)t(p ° + °+= =)t(p W)60t100cos(1600800 ° + + )60cos()20)(160( 2 1 )cos(IV 2 1 P ivmm °=θ−θ= =P W800 Chapter 11, Solution 2. First, transform the circuit to the frequency domain. °∠→ 030)t500cos(30 , 500 = ω 150jLjH3.0 =ω→ 100j- )10)(20)(500( j- Cj 1 F20 6- == ω →µ I I 1 I 2 + − 30∠0° V j 150 Ω - j 100 Ω 200 Ω 2.0j-902.0 150j 030 1 =°−∠= °∠ =I )t500sin(2.0)90t500cos(2.0)t(i 1 = ° − = 06.0j12.056.261342.0 j2 3.0 100j200 030 2 +=°∠= − = − °∠ =I )56.25t500cos(1342.0)t(i 2 ° + = ° ∠ = − =+= 49.4-1844.014.0j12.0 21 III )35t500cos(1844.0)t(i ° − = For the voltage source, ])35t500cos(1844.0[])t500cos(30[)t(i)t(v)t(p ° − × == At , s2t = )351000cos()1000cos(532.5p ° − = )935.0)(5624.0)(532.5(p = =p W91.2 For the inductor, ])t500sin(2.0[])t500cos(30[)t(i)t(v)t(p × == At , s2t = )1000sin()1000cos(6p = )8269.0)(5624.0)(6(p = =p W79.2 For the capacitor, ° ∠ == 63.44-42.13)100j-( 2c IV )56.25t500cos(1342.0[])44.63500cos(42.13[)t(i)t(v)t(p °+ × ° − == At , s2t = )56.261000cos()44.631000cos(18p ° + °−= )1329.0)(991.0)(18(p = =p W37.2 For the resistor, ° ∠ == 56.2584.26200 2R IV ])56.26t500cos(1342.0[])56.26t500cos(84.26[)t(i)t(v)t(p °+ × ° + == At , s2t = )56.251000(cos602.3p 2 °+= 2 1329.0)(602.3(p = =p W0636.0 Chapter 11, Solution 3. 10 , °∠→°+ 3010)30t2cos( 2 = ω 2jLjH1 =ω→ -j2 Cj 1 F25.0 = ω → 4 Ω 2 Ω I I 1 I 2 + − 10∠30° V j 2 Ω - j 2 Ω 2j2 2 )2j2)(2j( )2j2(||2j += − =− °∠= ++ °∠ = 565.11581.1 2j24 3010 I °∠=== 565.101581.1j 2 2j 1 III °∠= − = 565.56236.2 2 2j2 2 II For the source, )565.11-581.1)(3010( 2 1 * °∠°∠== IVS 5.2j5.718.43905.7 + = ° ∠ =S The average power supplied by the source = W5.7 For the 4-Ω resistor, the average power absorbed is === )4()581.1( 2 1 R 2 1 P 2 2 IW5 For the inductor, 5j)2j()236.2( 2 1 2 1 2 L 2 2 === ZIS The average power absorbed by the inductor = W0 For the 2-Ω resistor, the average power absorbed is === )2()581.1( 2 1 R 2 1 P 2 2 1 IW5.2 For the capacitor, 5.2j-)2j-()581.1( 2 1 2 1 2 c 2 1 === ZIS The average power absorbed by the capacitor = W0 Chapter 11, Solution 4. 20 Ω 10 Ω I 2 I 1 + − - j 10 Ω 50 V j 5 Ω For mesh 1, 21 10j)10j20(50 II + −= 21 j)j2(5 II +−= (1) For mesh 2, 12 10j)10j5j10(0 II + −+= 12 2j)j2(0 II +−= (2) In matrix form,             − − =       2 1 j22j jj2 0 5 I I 4j5 −=∆ , )j2(5 1 − = ∆ , -j10 2 = ∆ °∠= − − = ∆ ∆ = 1.12746.1 4j5 )j2(5 1 1 I °∠== ∆ ∆ = 66.128562.1 j4-5 j10- 2 2 I For the source, °∠== 12.1-65.43 2 1 * 1 IVS The average power supplied = °= )1.12cos(65.43 W68.42 For the 20-Ω resistor, == R 2 1 P 2 1 IW48.30 For the inductor and capacitor, =P W0 For the 10-Ω resistor, == R 2 1 P 2 2 IW2.12 Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 Ω 2 Ω + − j6 –j2 8∠–40˚ W4159.11 2 6828.1 P 38.256828.1 2j26j )2j2(6j 1 408 I 2 1 1 == °−∠= −+ − + °−∠ = Ω Ω P 3H = P 0.25F = 0 W097.52 2 258.2 P 258.238.256828.1 2j26j 6j I 2 2 2 == =°−∠ −+ = Ω Ω Chapter 11, Solution 6. 20 Ω 10 Ω I 2 I 1 + − - j 10 Ω 50 V j 5 Ω For mesh 1, 04)604(2j)2j4( o1 = + °∠−+ VI (1) )604(2 2o IV −°∠= (2) For mesh 2, 04)604(2)j2( o2 = − °∠−− VI (3) Substituting (2) into (3), 0)604(8608)j2( 22 = − ° ∠ − °∠−− II j10 6040 2 − °∠ =I Hence, j10 608j- j10 6040 6042 o − °∠ =       − °∠ −°∠=V Substituting this into (1),       − − °∠= − °∠ +°∠=+ j10 j14 )608j( j10 6032j 608j)2j4( 1 I °∠= + +°∠ = 125.06498.2 8j21 )14j1)(604( 1 I === )4()498.2( 2 1 R 2 1 P 2 2 14 IW48.12 Chapter 11, Solution 7. 20 Ω 10 Ω I 2 I 1 + − - j 10 Ω 50 V j 5 Ω Applying KVL to the left-hand side of the circuit, oo 1.04208 VI +=°∠ (1) Applying KCL to the right side of the circuit, 0 5j105j 8 11 o = − ++ VV I But, o11o 10 5j10 5j10 10 VVVV − =→ − = Hence, 0 1050j 5j10 8 o oo =+ − + V VI oo 025.0j VI = (2) Substituting (2) into (1), )j1(1.0208 o + =°∠ V j1 2080 o + °∠ =V °∠== 25- 2 10 10 o 1 V I =             == )10( 2 100 2 1 R 2 1 P 2 1 IW250 Chapter 11, Solution 8. We apply nodal analysis to the following circuit. At node 1, I o V 2 V 1 6∠0° A 0.5 I o j 10 Ω - j 20 Ω I 2 40 Ω 20j-10j 6 211 VVV − += 21 120j VV − = (1) At node 2, 40 5.0 2 oo V II =+ But, j20- 21 o VV I − = Hence, 40j20- )(5.1 221 VVV = − 21 )j3(3 VV −= (2) Substituting (1) into (2), 0j33360j 222 = + − − VVV j6)-1( 37 360 j6 360j 2 += − =V j6)-1( 37 9 40 2 2 +== V I =       == )40( 37 9 2 1 R 2 1 P 2 2 2 I W78.43 Chapter 11, Solution 9. rmsV8)2)(4(V 2 6 1V so ==       += === mW 10 64 R V P 2 o 10 mW4.6 The current through the 2 -kΩ resistor is mA1 k2 V s = == RIP 2 2 mW2 Similarly, == RIP 2 6 mW6 Chapter 11, Solution 10. No current flows through each of the resistors. Hence, for each resistor, =P W0 . Chapter 11, Solution 11. , , 377=ω 4 10R = -9 10200C ×= 754.0)10200)(10)(377(RC -94 =×=ω °=ω 02.37)RC(tan -1 Ω°∠=°∠ + = k37.02-375.637.02- )754.0(1 k10 Z 2 ab mA)68t377cos(2)22t377sin(2)t(i ° − = °+= °∠= 68-2I 3 2 -3 ab 2 rms 10)37.02-375.6( 2 102 ZIS ×°∠       × == mVA37.02-751.12S °∠= == )02.37cos(SP mW181.10 Chapter 11, Solution 12. (a) We find using the circuit in Fig. (a). Th Z Z th 8 Ω - j 2 Ω (a) 882.1j471.0)4j1( 17 8 j28 (8)(-j2) -j2||8 Th −=−= − ==Z == * ThL ZZ Ω + 882.1j471.0 We find using the circuit in Fig. (b). Th V I o + V th − - j 2 Ω 4 ∠ 0° A 8 Ω (b) )04( 2j8 2j- o °∠ − =I j28 j64- I8 oTh − ==V =       == )471.0)(8( 68 64 R8 P 2 L 2 Th max V W99.15 (b) We obtain from the circuit in Fig. (c). Th Z 5 Ω - j 3 Ω Z th j 2 Ω 4 Ω (c) 167.1j5.2 3j9 )3j4)(5( 2j)3j4(||52j Th += − − +=−+=Z == * ThL ZZ Ω − 167.1j5.2