已知正弦交流电流i1=10sin(at+45°)A,i2=10sin(at-45°)A,则i=i1+i2=?A.
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10A这么算的:i1最大值为10A,因此有效值为10/√2.i2同理为10/√2.
则i1=10/√2∠45° i2=10/√2∠-45°.
因此i1+i2=10/√2∠45°+10/√2∠-45°=5+j5+5-j5=10A
======以下答案可供参考======
供参考答案1:
i1=10sin(at+45°),i2=10sin(at-45°)
i=i1+i2=10sin(at+45°)+10sin(at-45°)
=10(sinat*√2/2+cosat*√2/2)+10(sinat*√2/2-cosat*√2/2)
=10√2sinat