Answer: 62 cos(50t - 70°)
Step-by-step explanation:
First we need to convert all sines into cosines because phasor forms are represented through cosine. For that we will use the fact that sin(wt + ∅) = cos(wt + ∅ - 90°)
Therefore, 37sin50t = 37cos(50t - 90°)
Now we have 37cos(50t - 90°)+ 30 cos(50t – 45°). We need to convert them into phasor form to add the terms. For that we will use the fact Acos(wt+∅)=A∠∅ which can be represented using real and imaginary parts as A [cos(∅)+jsin(∅)].
So,
37cos(50t - 90°)
= 37∠-90°
= 37[cos(-90°)+jsin(-90°)
=37[0+j(-1)]
= -j37
Similarly,
30 cos(50t – 45°)
=30∠-45
=30[cos(-45)+jsin(-45)
=30[0.707-j0.707]
=21.21 - j21.21
37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = 21.21 - j58.21
Now we need to convert the real and imaginary parts back to cosine form. We will first calculate the magnitude by the formula √a²+b² where a and b are the real and imaginary parts respectively.
Here a=21.21 and b=58.21
magnitude = √(21.21)²+(58.21)²=61.95≅62
For calculating phase ∅ the formula is ∅=inversetan (b/a) where a and b are the real and imaginary parts respectively.
∅=inversetan(-58.21/21.21)
= -69.9°≅-70°
So the final answer is 62cos(50t-70°)