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answer10+12+13

answer10+12+13
answer10+12+13

第十、十二章计算题答案

10-1 略

10-2 (a )2/3H (b ) 4H (c )0.5H

10-3 Ω∠= 613.3in Z

10-4 A ∠=m 663.31

I V 6622

∠=U 10-5 当20=n 时,L R 获得的最大功率, W 069.0max =P

12-1 设V 0220A

∠=U , A 2.5214.3A

-∠=I 对应可得到B 、C 相 =''B A U 93.308.369

∠=V ,对应可得到B 、C 相 12-2 设V 0220A

∠=U ⑴ A 13.5311.3A -∠=I A 13.17311B

-∠=I A 87.6611C ∠=I 13.53cos 113803cos 3L

L ???==?I U P Y W 983.4343= ⑵ 线电流 A 13.5333A -∠=I A 13.17333B

-∠=I A 87.6633C ∠=I 相电流A 13.2305.19A B -∠=I A 13.14305.19BC -∠=I A 87.9605.19C

∠=I 13.53cos 333803cos 3L L ???==??I U P W 95.13031=

⑶ PY P Δ3I I = LY L Δ3I I = PY P Δ3U U = LY L ΔU U = Y Δ3P P = 12-3 有中线时

A 3022 -∠=A I A 12022

B ∠=I A 18022 ∠=C

I C

B A N I I I I ++=A 150105.16 ∠= 无中线时 ()V 526.80j 474.139150051.161

'+-=∠= N N U A 30105.38 -∠=A I A 165725.19 ∠=B I C

I A 135725.19 ∠= 12-4 1P 951.9W = 8W .19032=P

13- A )]776.292000cos(2724.0)29.31000cos(2719.4[ -+-=t t i ,

A 774.4=I

W 91.227=P

13-2 21i i i +=A )]11.24cos(2045.2110[ -+=t ω,

A 3.23=I

W 122.3117=P

13-3 A )]762.61cos(2121.25[1 -+=t i ω ()A 072.118c o s 2588.02 -=t i ω

13-4 V 944.848221=+=U V 4042=+=

U

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