The size conductor that should be used to supply each motor is 12 AWG and the size conductor that should be used for the feeder conductor is 1/0AWG.
<h3>Size conductor</h3>
a. Using NEC table 430.250
Running connect for 150HP connected to 460V = 21 A
Conductor size=25% current to 125% current (Increase)
Increase current=21×125/100
Increase current=26.25 A
Nearest standard conductor size using NEC table 310.15(B)(16)=25 A
Since the motor contain NEMA design code the conductor will be choosing from 75° column which means that the size conductor that should be used to supply each motor is 12AWG copper conductors.
b. Feeder conductor
Using NEC section 430.24 plus ampere rating of other motors
Running current
Using NEC Table 430.250
Running current=26.25+21+21+21+21+21
Running current=131.25 A
The value that is greater than 131.25 A under 75° column of NEC Table 310 (B)(16) is 150 A which means that the size conductor that should be used for the feeder conductor is 1/0 AWG conductors.
Therefore the size conductor that should be used to supply each motor is 12 AWG and the size conductor that should be used for the feeder conductor is 1/0AWG.
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Solution:
Given:
Calcium and Oxygen ions each having valence = 2 or 2 valence electrons
seperation distance, r = 1.25nm = 
m
Coulombian force for two point charges seperated by a distance 'r' is given by:
F = 
(1)
Now, we know that
q = ne
where, n = no. of electrons
e = charge of an electron = 
C
= 
e = 2e
= 
e = 2e
Using Eqn (1)
F = 
F = 
N
Since, the force is between two opposite charged ions, it is attractive in nature.
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