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earnstyle [38]
2 years ago
10

3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?

Physics
1 answer:
ZanzabumX [31]2 years ago
5 0

Answer:

E_s = 277.13V

Explanation:

Given

Load\ Voltage = 480V

Required

Determine the voltage dropped in each stage.

The relation between the load voltage and the voltage dropped in each stage is

E_l = E_s * \sqrt3

Where

E_l  = 480

So, we have:

480 = E_s * \sqrt3

Solve for E_s

E_s = \frac{480}{\sqrt3}

E_s = \frac{480}{1.73205080757}

E_s = 277.128129211

E_s = 277.13V

<em>Hence;</em>

<em>The voltage dropped at each phase is approximately 277.13V</em>

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According to Newton's law of universal gravitation:

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F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

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