In a 480—240-volt 3-wire transformer delivering power to a 3Ø load, what is the current flowing in each circuit conductor if the coil current is 75 amps?

In a 480—240-volt 3-wire transformer supplying power to a three-phase load, the relationship between the primary side (480 volts) and the secondary side (240 volts) is governed by the transformer's turn ratio. In this scenario, the transformer likely steps down the voltage from 480 volts to 240 volts, meaning that the current on the secondary side will increase in proportion to the decrease in voltage, according to the principle of conservation of power.

When current flows through the transformer, the equation P = V x I (Power equals Voltage times Current) helps us understand that as the voltage drops, the current must rise to maintain equivalent power on the secondary side, assuming ideal operating conditions with no losses.

Given that the coil (primary) current is 75 amps on the 480-volt side, the power on that side can be calculated as follows:

[ P_{primary} = V_{primary} \times I_{primary} = 480 , \text{V} \times 75 , \text{A} = 36,000 , \text{W} ]

For the secondary side, that power must remain the same under ideal conditions:

[ P_{secondary} = V