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fgiga [73]
3 years ago
5

Using phasors, the value of 37 sin 50t + 30 cos(50t – 45°) is _________ cos(50t+(_____°)). Please report your answer so the magn

itude is positive and all angles are in the range of negative 180 degrees to positive 80 degrees
Engineering
1 answer:
Mars2501 [29]3 years ago
7 0

Answer: 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 = <u>37cos(50t - 90°)</u>

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)]

= <u>-j37</u>

Similarly,

30 cos(50t – 45°)

=30∠-45

=30[cos(-45)+jsin(-45)

=30[0.707-j0.707]

=<u>21.21 - j21.21</u>

37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = <u>21.21 - j58.21</u>

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≅<u>62</u>

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°≅<u>-70°</u>

So the final answer is 62cos(50t-70°)

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Answer:

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Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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