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What is the maximum safe secondary current for a 40 VA transformer that has a primary voltage of 240 V and a secondary voltage of 24 V?

Select one:
A. [tex]\frac{240 V}{24 V} = 10 A[/tex]
B. [tex]\frac{240 V}{40 VA} = 6 A[/tex]
C. [tex]\frac{24 V}{40 VA} = 0.6 A[/tex]
D. [tex]\frac{40 VA}{24 V} = 1.67 A[/tex]


Sagot :

To determine the maximum safe secondary current for the given transformer, we need to use the formula:

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

In this scenario, we have:
- Power (VA) = 40 VA (Volt-Amperes)
- Secondary Voltage (V) = 24 V

Substitute the given values into the formula:

[tex]\[ \text{Current} = \frac{40 \, \text{VA}}{24 \, \text{V}} \][/tex]

Performing this division:

[tex]\[ \text{Current} = 1.67 \, \text{A} \][/tex]

Thus, the maximum safe secondary current for the transformer is [tex]\(1.67 \, \text{A}\)[/tex].

The correct answer is:

D. [tex]\(40 \, \text{VA} \div 24 \, \text{V} = 1.67 \, \text{A}\)[/tex]