已知u1=10sin(314t+π/2)V

更新时间：2022-05-08 00:43:27 81次访问

已知u1=10sin(314t+π/2)V，u2=10sin(314t+π/3)V，试分别用相量求：
1）u1 + u2
2） u1－u2

u₁+u₂

=10[sin(314t+π/2)+sin(314t+π/3)]

=10*2sin[314t+(π/2+π/3)/2]cos[π/2-π/3)/2]

=20cos(π/12)sin(314t+5π/12)

=5(√6+√2)sin(314t+5π/12)

u₁-u₂

=10[sin(314t+π/2)-sin(314t+π/3)]

=10*2cos[314t+(π/2+π/3)/2]sin[(π/2-π/3)/2]

=20sin(π/12)cos(314t+5π/12)

=5(√6-√2)sin(314t+11π/12)

希望以下回答对您有帮助：u₁+u₂
=10[sin(314t+π/2)+sin(314t+π/3)]
=10*2sin[314t+(π/2+π/3)/2]cos[π/2-π/3)/2]
=20cos(π/12)sin(314t+5π/12)
=5(√6+√2)sin(314t+5π/12)u₁-u₂
=10[sin(314t+π/2)-sin(314t+π/3)]
=10*2cos[314t+(π/2+π/3)/2]sin[(π/2-π/3)/2]
=20sin(π/12)cos(314t+5π/12)
=5(√6-√2)sin(314t+11π/12)