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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)?

User Novicef
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1 Answer

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11 votes

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


(I_p)/(I_s)=(V_s)/(V_p)

Where:


\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}

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


I_p=I_s(V_s)/(V_p)

Now, we plug in the values:


I_p=(355mA)((12V)/(120V))

Solving the operations:


I_p=35.5mA

Therefore, the output current is 35.5 mA.

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


P=VI

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


P=(0.355A)(120V)

Solving the operations:


P=42.6W

Therefore, the power is 42.6 Watts.

User Mbethke
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