A single-phase 240-volt load drawing 43 amps is connected to a transformer with a primary voltage of 480 volts. How much current is flowing in the primary winding?

To find the current flowing in the primary winding of the transformer, you need to apply the principle of conservation of power, which states that power in the primary winding equals power in the secondary winding (ignoring losses for simplicity).

In this scenario, you have a single-phase system, so the relationship between primary voltage, primary current, secondary voltage, and secondary current can be expressed using the formula:

[ P_{primary} = P_{secondary} ]

This can be simplified using the relationship:

[ V_{primary} \times I_{primary} = V_{secondary} \times I_{secondary} ]

Given that the secondary load draws 43 amps at 240 volts, you can calculate the power in the secondary:

[ P_{secondary} = 240 \text{ volts} \times 43 \text{ amps} = 10,320 \text{ watts} ]

Now, to relate this to the primary side, we know the primary voltage is 480 volts. Rearranging the power equation to solve for the primary current gives us:

[ I_{primary} = \frac{P_{secondary}}{V_{primary}} ]

Substituting the values you have:

[ I_{primary} =