A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 355 mA.(a) What is the current input (in mA)?(b) What is the power input (in W)?

Respuesta :

We are asked to determine the current input for a transformer. To do that we will use the following formula:

[tex]\frac{I_p}{I_s}=\frac{V_s}{V_p}[/tex]

Where:

[tex]\begin{gathered} I_p=\text{ primary or input current} \\ I_s=\text{ secondary or output current} \\ V_s=\text{ secondary voltage} \\ V_p=\text{ primary voltage} \end{gathered}[/tex]

Now, we solve for the primary current by multiplying both sides by the secondary current:

[tex]I_p=I_s\frac{V_s}{V_p}[/tex]

Now, we plug in the values:

[tex]I_p=(355mA)(\frac{12V}{120V})[/tex]

Solving the operations:

[tex]I_p=35.5mA[/tex]

Therefore, the output current is 35.5 mA.

Part b. To determine the power input we use the following formula:

[tex]P=VI[/tex]

Since we want to determine the input power we use the value of primary current and voltage:

[tex]P=(0.355A)(120V)[/tex]

Solving the operations:

[tex]P=42.6W[/tex]

Therefore, the power is 42.6 Watts.