What is the maximum current draw of a load connected to a step down transformer rated at 40 VA with a primary voltage of 120V and secondary voltage of 24V?

What is the maximum current draw of a load connected to a step down transformer rated at 40 VA with a primary voltage of 120V and secondary voltage of 24V?

• A. 1.5 amps
• B. 1.7 amps
• C. 2.0 amps
• D. 2.5 amps

Answer: (B)

To determine the maximum current draw on the secondary side of the transformer, we can use the transformer formulas pertaining to voltage and current. A transformer's power rating, expressed in volt-amperes (VA), indicates how much current can be drawn by the load at the specified secondary voltage. The relationship between power (in VA), voltage, and current is given by the formula: \[ \text{Power (VA)} = \text{Voltage (V)} \times \text{Current (A)} \] For this transformer rated at 40 VA with a secondary voltage of 24V, we can rearrange the formula to find the maximum current draw on the secondary side: \[ \text{Current (A)} = \frac{\text{Power (VA)}}{\text{Voltage (V)}} \] Plugging in the values: \[ \text{Current (A)} = \frac{40 \text{ VA}}{24 \text{ V}} \] Calculating this gives: \[ \text{Current (A)} = 1.67 \text{ A} \] Rounding to one decimal place leads to the conclusion that the maximum current draw is approximately 1.7 amps. This reasoning aligns with the correct

To determine the maximum current draw on the secondary side of the transformer, we can use the transformer formulas pertaining to voltage and current. A transformer's power rating, expressed in volt-amperes (VA), indicates how much current can be drawn by the load at the specified secondary voltage.

The relationship between power (in VA), voltage, and current is given by the formula:

[ \text{Power (VA)} = \text{Voltage (V)} \times \text{Current (A)} ]

For this transformer rated at 40 VA with a secondary voltage of 24V, we can rearrange the formula to find the maximum current draw on the secondary side:

[ \text{Current (A)} = \frac{\text{Power (VA)}}{\text{Voltage (V)}} ]

Plugging in the values:

[ \text{Current (A)} = \frac{40 \text{ VA}}{24 \text{ V}} ]

Calculating this gives:

[ \text{Current (A)} = 1.67 \text{ A} ]

Rounding to one decimal place leads to the conclusion that the maximum current draw is approximately 1.7 amps.

This reasoning aligns with the correct