ashk537 ashk537

14-02-2020
Physics

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For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 20.0 A for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current

Respuesta :

oodaramola oodaramola

17-02-2020

Answer:

P=I^2*R=400R where R is the resistance of the wire

Explanation:

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