已知Um=311V,ω=314rad/s,φ=-30°。则正弦电压的解析式为( )
A. u=311sin(314t+30°)
B. u=311sin(314t-30°)
C. u=311√2sin(314t+30°)
D. u=311√2sin(314t-30°)
已知u1=100sin(628t+45°),u2=141sin(628t-30°),则这两个电压的相位差是()
A. 15°
B. 45°
C. 30°
D. 75°
指出i = 10sin(314 t + 20o)A的振幅是( )A、角频率是( )rad/s 、频率是( )Hz、初相位是( )o