题目

已知电流I=3sin(ωt),电压U=4sin(ωt+)(t＞0),且电功率P=IU,则电功率P的最大值是（    ）A.5                  B.7                 C.6                  D.12 答案：答案:C解析:P=IU=3sin(ωt)×［4sin(ωt+)］=3sin(ωt)·4cos(ωt)=6sin(2ωt).∴电功率P的最大值是6.

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