A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the types of transformer being used.

Options

A. `0.27 mA`
B. `2.7 A`
C. `0.27 A`
D. `10 A`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Power in the secondary (PS) = 60 W
- Current in the secondary (IS) = 0.54 A
- Voltage in the primary (VP) = 220 V (standard line voltage)

### Step 2: Calculate the voltage across the secondary (VS)
Using the formula for power in the secondary:
\[ PS = VS \times IS \]
We can rearrange this to find VS:
\[ VS = \frac{PS}{IS} \]
Substituting the known values:
\[ VS = \frac{60 \, \text{W}}{0.54 \, \text{A}} \]
Calculating this gives:
\[ VS \approx 111.11 \, \text{V} \]

### Step 3: Determine the type of transformer
Since the voltage across the primary (VP = 220 V) is greater than the voltage across the secondary (VS ≈ 111.11 V), we conclude that this is a **step-down transformer**. In a step-down transformer, the primary voltage is higher than the secondary voltage.

### Step 4: Use the transformer relationship to find the primary current (IP)
The relationship between the primary and secondary currents and voltages in a transformer is given by:
\[ \frac{VS}{VP} = \frac{IP}{IS} \]
Rearranging this to find IP gives:
\[ IP = IS \times \frac{VS}{VP} \]
Substituting the known values:
\[ IP = 0.54 \, \text{A} \times \frac{111.11 \, \text{V}}{220 \, \text{V}} \]
Calculating this gives:
\[ IP \approx 0.27 \, \text{A} \]

### Conclusion
The current in the primary coil (IP) is approximately **0.27 A**. The transformer used is a **step-down transformer** because the primary voltage is greater than the secondary voltage.

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