Respuesta :
Answer:
I = 14.44A
Explanation:
calculating 80% of the circuit breaker current
I = (80/100)20A = 16A
eight 500W lamps are to be installed on the circuit. assume they are connected in parallel
total power = 500 x 8 = 4000W
power = voltage x current
current = power/voltage = 4000W/277V = 14.4A
The current obtained is less than 80% of the circuit breaker.
20A circuit breaker is large enough to carry the load.
The watt is used to quantify the amount of power in a circuit
Yes the circuit is large enough to carry the load
The reason the above response is correct is as follows:
The given parameters are:
Number of 500-W lamp to be installed on the same circuit = 8
Circuit voltage, V = 227 V
Rating of circuit breaker on the circuit = 20-A
Allowable load for continuous use = 80% of the rating
Required:
To find the capacity of the circuit to carry the load
Solution:
The total power available in the circuit, [tex]P_{tot}[/tex] = 500-W × 8 = 4,000-W
Power, P = Voltage, V × Current, I
[tex]I = \dfrac{Power}{Voltage}[/tex]
The total current required in the circuit, [tex]I_{tot}[/tex], is therefore;
[tex]I_{tot} = \dfrac{P_{tot}}{V}[/tex]
[tex]I_{tot} = \dfrac{4,000\, W}{277 \, V} \approx 14.44 \, A[/tex]
The total current flowing in the circuit, [tex]I_{tot}[/tex] ≈ 14.44 A
The allowable continuous use current in the circuit, [tex]I_{allowabe}[/tex] = 80% of 20-A = 16-A
Therefore, the allowable current load of the circuit is more than the total current flowing in the circuit, [tex]I_{allowabe}[/tex] > [tex]I_{tot}[/tex] and the circuit is large enough to carry the load
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