- #1
vvl92
- 13
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I am given the problem:
100Sin(ωt+30) +20cos(ωt)
and the solution is 78.7<38.9degrees
How do I convert to phasor form?
I know for this calculation I need the following relationship:
√2Esin(ωt-∅)=Ee^(-j∅)
At first I tried making both sines and then splitting up 100Sin(ωt+30) and 20Sin(ωt+90)using the trigonometric identities but I get stuck at that. Please help!
100Sin(ωt+30) +20cos(ωt)
and the solution is 78.7<38.9degrees
How do I convert to phasor form?
I know for this calculation I need the following relationship:
√2Esin(ωt-∅)=Ee^(-j∅)
At first I tried making both sines and then splitting up 100Sin(ωt+30) and 20Sin(ωt+90)using the trigonometric identities but I get stuck at that. Please help!
