Answer:
[tex]vs(t)=152.08cos(\omega t -45)=152.08\angle-45[/tex]
Step-by-step explanation:
A tension expressed in form:
[tex]V(t)=Vm*cos(\omega t+\phi)[/tex]
It can be converted directly into a fasor:
[tex]V(t)=Vm\angle \phi[/tex]
If the function is a positive sine we subtract 90 °
[tex]V(t)=Vm*sin(\omega t+\phi)=Vm*cos(\omega t+\phi-90)[/tex]
So:
[tex]v1(t)=100cos(\omega t)= 100\angle0\\v2(t)=100sin(\omega t)=100cos(\omega t-\ph90)=100\angle -90[/tex]
[tex]vs(t)=v1(t)+v2(t)=100\angle 0 +100\angle-90=152.0811931\angle-45\approx152.08\angle-45[/tex]
Therefore:
[tex]vs(t)=152.08\angle-45=152.08cos(\omega t -45)[/tex]
