1cos314t 3v i 1414 cos314t 6 a b1000hz i1

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Unformatted text preview: 解:上式 = 180.2 + j126.2 + 20.62∠14.04o = 180.2 + j126.2 + 6.728∠70.16o = 180.2 + j126.2 + 2.238 + j 6.329 = 182.5 + j132.5 = 225.5∠36o 5 §8-2 例 3: 相量法的基本概念 已知 i1 = 12 .7 2 cos (314 t + 30 ° )A i 2 = 11 2 c os (314 t − 60 ° )A 求:i = i1 + i2 。 解: 用相量计算 & I 1 = 12.7 30 ° A & I 2 = 11 − 60 ° A &&& I = I 1 + I 2 = 12.7 30° + 11 − 60° = 12.7( 30 ° + jsin 30 °) +11(cos60 ° − jsin 60 °) cos = (16.5 - j3.18)A = 16.8 − 10.9 ° A i = 16....
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