Answer:
The three phase full load secondary amperage is 2775.7 A
Explanation:
Following data is given,
S = Apparent Power = 1000 kVA
No. of phases = 3
Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>
<em>Since,</em>
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![V_{1ph} = 120 V](https://tex.z-dn.net/?f=V_%7B1ph%7D%20%3D%20120%20V)
The formula for apparent power in three phase system is given as:
![S = \sqrt{3} VI](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%7B3%7D%20VI)
Where:
S = Apparent Power
V = Line Voltage
I = Line Current
In order to calculate the Current on Secondary Side, substituting values in above formula,
![1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A](https://tex.z-dn.net/?f=1000%20kVA%20%3D%20%5Csqrt%7B3%7D%20%2A%20%28208%29%20%2A%20%28I%29%5C%5C1000%20%2A%201000%20%3D%20%5Csqrt%7B3%7D%20%2A%20%28208%29%20%2A%20%28I%29%5C%5CI%20%3D%20%5Cfrac%7B1000%20%2A%201000%7D%7B%5Csqrt%7B3%7D%20%2A%20%28208%29%20%7D%5C%5C%20I%20%3D%202775.7%20A)
Answer:
When the brakes are applied the in the typical double transverse wishbone front suspension, it "drives" the car ground due to the setting of the link-type system pivot points on the lower wishbone are have parallel alignment to the road
Explanation:
In order to minimize the car's reaction to the application of the brakes, the front and rear pivot are arranged with the lower wishbone's rear pivot made to be higher than the front pivot as such the inclined wishbone torque results in an opposing vertical force to the transferred extra weight from the back due to breaking.