A transformer of efficiency 90% draws an input power of 4 kW. An electrical appliance connected across the secondary draws a current of 6A. The impedance of the device is

Options

A. `60Omega`
B. `50Omega`
C. `80Omega`
D. `100Omega`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Efficiency of the transformer (η) = 90% = 0.9
- Input power (P_in) = 4 kW = 4000 W
- Current drawn by the appliance (I) = 6 A

### Step 2: Calculate the output power (P_out)
The efficiency of a transformer is given by the formula:
\[ \text{Efficiency} (\eta) = \frac{P_{\text{out}}}{P_{\text{in}}} \]
From this, we can rearrange the formula to find the output power:
\[ P_{\text{out}} = \eta \times P_{\text{in}} \]
Substituting the values:
\[ P_{\text{out}} = 0.9 \times 4000 \, \text{W} = 3600 \, \text{W} = 3.6 \, \text{kW} \]

### Step 3: Use the power formula to find impedance
The power consumed by the appliance can also be expressed as:
\[ P = I^2 \times Z \]
Where:
- \( P \) is the output power (3.6 kW or 3600 W)
- \( I \) is the current (6 A)
- \( Z \) is the impedance

Rearranging the formula to solve for impedance (Z):
\[ Z = \frac{P}{I^2} \]

### Step 4: Substitute the known values into the impedance formula
Substituting the values we have:
\[ Z = \frac{3600 \, \text{W}}{(6 \, \text{A})^2} \]
Calculating \( (6 \, \text{A})^2 \):
\[ (6 \, \text{A})^2 = 36 \, \text{A}^2 \]
Now substituting back:
\[ Z = \frac{3600 \, \text{W}}{36 \, \text{A}^2} = 100 \, \Omega \]

### Final Answer
The impedance of the device is \( 100 \, \Omega \).

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