Answer :
Final answer
This is calculated using the formula [tex]( I = \frac{P}{\sqrt{3} \times V \times \text{Power Factor}} \),[/tex] where [tex]\( P \)[/tex]is the power in watts, [tex]\( V \)[/tex] is the voltage, and the power factor is typically 0.8 for 3-phase motors.
The correct option is b. 31.7 amps
Explanation
To find the amperage drawn by the motor, we can use the formula:
[tex]\[ I = \frac{P}{\sqrt{3} \times V \times \text{Power Factor}} \][/tex]
where:
[tex]\( I \)[/tex] = Current (in amps)
[tex]\( P \)[/tex]= Power (in watts) = 10 hp [tex]\( \times \) 746 watts/hp[/tex]
[tex]\( V \)[/tex] = Voltage (in volts) = 208 volts
For 3-phase motors, the power factor is typically around 0.8.
Substituting the given values, we get:
[tex]\[ I = \frac{10 \times 746}{\sqrt{3} \times 208 \times 0.8} \][/tex]
[tex]\[ I ≈ 31.7 \text{ amps} \][/tex]
Therefore, the correct answer is b. 31.7 amps.
The formula[tex]\( I = \frac{P}{\sqrt{3} \times V \times \text{Power Factor}} \)[/tex]is derived from the power formula [tex]\( P = VI\cos(\theta) \), where \( \theta \)[/tex] is the phase angle. For balanced 3-phase systems, [tex]\( \cos(\theta) \)[/tex] is typically equal to the power factor. We divide by [tex]\( \sqrt{3} \)[/tex]because it's a 3-phase system, and we use the rms (root mean square) values of voltage and current.
Given the power factor of 0.8, which is common for induction motors, we account for the phase difference between voltage and current. This ensures that the calculated current represents the effective current in the system. Multiplying the power in horsepower by 746 converts it to watts, which is the unit consistent with the formula for current calculation.
Finally, plugging the values into the formula yields the amperage drawn by the motor, which is approximately 31.7 amps. This calculation helps determine the electrical load and assists in proper sizing of protective devices and conductors for the motor circuit.