AWG #10 copper on a 30 amp breaker.
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no
It would be at least 250 amps, maybe 300 amps.
The formula for amps is I = W/E. Amps = 40/240 = .17 primary amperage. For the secondary amperage I = W/E. Amps = 40/24 = 1.7 amps.
10A
For a single-phase induction motor, allow 7 amps on a 240 v for a 1-HP motor. Therefore the formula is: current = 7 X HP x 240 / voltage