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Sagot :
To determine how much additional current this motor can safely handle for intermittent power demands, follow these steps:
1. Identify the Full Load Current (FLC):
The motor has a Full Load Current (FLC) of 10 Amps.
2. Identify the Service Factor (SF):
The Service Factor (SF) of the motor is 1.15. This factor indicates that the motor can handle up to 1.15 times its full load current under certain conditions.
3. Calculate the Total Current:
The total current that can be handled by the motor is the FLC multiplied by the SF.
[tex]\[ \text{Total Current} = \text{FLC} \times \text{SF} \][/tex]
Substituting the given values:
[tex]\[ \text{Total Current} = 10 \text{ A} \times 1.15 = 11.5 \text{ A} \][/tex]
4. Calculate the Additional Current:
The additional current is the difference between the total current and the full load current:
[tex]\[ \text{Additional Current} = \text{Total Current} - \text{FLC} \][/tex]
Substituting the values:
[tex]\[ \text{Additional Current} = 11.5 \text{ A} - 10 \text{ A} = 1.5 \text{ A} \][/tex]
So, the motor can handle an additional 1.5 A safely for intermittent power demands.
Thus, the correct answer is:
a) [tex]$1.5 \text{ A}$[/tex]
1. Identify the Full Load Current (FLC):
The motor has a Full Load Current (FLC) of 10 Amps.
2. Identify the Service Factor (SF):
The Service Factor (SF) of the motor is 1.15. This factor indicates that the motor can handle up to 1.15 times its full load current under certain conditions.
3. Calculate the Total Current:
The total current that can be handled by the motor is the FLC multiplied by the SF.
[tex]\[ \text{Total Current} = \text{FLC} \times \text{SF} \][/tex]
Substituting the given values:
[tex]\[ \text{Total Current} = 10 \text{ A} \times 1.15 = 11.5 \text{ A} \][/tex]
4. Calculate the Additional Current:
The additional current is the difference between the total current and the full load current:
[tex]\[ \text{Additional Current} = \text{Total Current} - \text{FLC} \][/tex]
Substituting the values:
[tex]\[ \text{Additional Current} = 11.5 \text{ A} - 10 \text{ A} = 1.5 \text{ A} \][/tex]
So, the motor can handle an additional 1.5 A safely for intermittent power demands.
Thus, the correct answer is:
a) [tex]$1.5 \text{ A}$[/tex]
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