A transformer has an efficiency of 80%. It works at 4 kW and 100 V. If secondary voltage is 240 V, the current in primary coil is

Options

A. `10 A`
B. `4 A`
C. `0.4` A
D. `40 A `

Answer

To solve the problem, we need to find the current in the primary coil of the transformer given its efficiency, input power, primary voltage, and secondary voltage.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Efficiency (η) = 80% = 0.8
- Input Power (P_in) = 4 kW = 4000 W
- Primary Voltage (V_p) = 100 V
- Secondary Voltage (V_s) = 240 V

2. **Calculate the Output Power (P_out)**:
The efficiency of a transformer is given by the formula:
\[ \text{Efficiency} (\eta) = \frac{P_{out}}{P_{in}} \]
Rearranging this gives:
\[ P_{out} = \eta \times P_{in} \]
Substituting the values:
\[ P_{out} = 0.8 \times 4000 = 3200 \text{ W} \]

3. **Use the Power Formula to Find Current in the Primary Coil (I_p)**:
The power in the primary coil can also be expressed as:
\[ P_{in} = V_p \times I_p \]
Rearranging to find \( I_p \):
\[ I_p = \frac{P_{in}}{V_p} \]
Substituting the known values:
\[ I_p = \frac{4000 \text{ W}}{100 \text{ V}} = 40 \text{ A} \]

4. **Conclusion**:
The current in the primary coil is \( I_p = 40 \text{ A} \).

### Final Answer:
The current in the primary coil is **40 A**.