Using phasors, the value of 37 sin 50t + 30 cos(50t – 45°) is cos(50t+° report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 80 degrees

62 cos(50t - 70°)

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°)

answer:

an example of solubility is salt or sugar dissolving in water.

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