A `220 V` input is supplied to a transformer. The output circuit draws a current of `2.0 A` at `440 V`. If the efficiency of the transformer is `80%`, the current drawn by the primery winding of the transformer is

Options

A. `3.6 A`
B. `2.8 A`
C. `2.5 A`
D. `5.0 A`

Explanation

To solve the problem step by step, we will use the concept of transformer efficiency and the relationship between input and output power.

### Step 1: Understand the given values
- Input Voltage (Vp) = 220 V
- Output Voltage (Vs) = 440 V
- Output Current (Is) = 2.0 A
- Efficiency (η) = 80% = 0.80

### Step 2: Write the formula for efficiency
The efficiency of a transformer is given by the formula:
\[ η = \frac{P_{out}}{P_{in}} \times 100 \]
Where:
- \(P_{out}\) = Output Power
- \(P_{in}\) = Input Power

### Step 3: Calculate the output power
The output power can be calculated using the output voltage and current:
\[ P_{out} = V_s \times I_s \]
Substituting the values:
\[ P_{out} = 440 \, \text{V} \times 2.0 \, \text{A} = 880 \, \text{W} \]

### Step 4: Use the efficiency formula to find input power
Rearranging the efficiency formula to find input power:
\[ P_{in} = \frac{P_{out}}{η} \]
Substituting the values:
\[ P_{in} = \frac{880 \, \text{W}}{0.80} = 1100 \, \text{W} \]

### Step 5: Write the formula for input power
The input power can also be expressed as:
\[ P_{in} = V_p \times I_p \]
Where \(I_p\) is the current drawn by the primary winding. Rearranging gives:
\[ I_p = \frac{P_{in}}{V_p} \]

### Step 6: Substitute the known values to find \(I_p\)
Substituting the values we have:
\[ I_p = \frac{1100 \, \text{W}}{220 \, \text{V}} = 5 \, \text{A} \]

### Conclusion
The current drawn by the primary winding of the transformer is \(5 \, \text{A}\).

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