A power transmission line feeds input power at 2400V to a step-down transformer and which delivers power at 240V with its primary windings having 5000 turns . If the current in the primary coil of the transformer is 5A and its efficiency is 80%, then what is the output current (in A)?

Answer

To solve the problem, we need to find the output current of a step-down transformer given the input voltage, output voltage, primary current, and efficiency.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Input Voltage (\( V_p \)) = 2400 V
- Output Voltage (\( V_s \)) = 240 V
- Primary Current (\( I_p \)) = 5 A
- Efficiency (\( \eta \)) = 80% = 0.8

2. **Understand the Efficiency Formula**:
The efficiency of a transformer is given by the formula:
\[ \eta = \frac{P_{out}}{P_{in}} \]
where \( P_{out} \) is the output power and \( P_{in} \) is the input power.

3. **Express Power in Terms of Voltage and Current**:
Power can be expressed as:
\[ P = V \times I \]
Therefore, we can write:
\[ P_{in} = V_p \times I_p \]
\[ P_{out} = V_s \times I_s \]

4. **Substitute Power into the Efficiency Formula**:
Substituting the expressions for power into the efficiency formula gives:
\[ \eta = \frac{V_s \times I_s}{V_p \times I_p} \]

5. **Rearranging for Output Current**:
Rearranging the equation to solve for the output current (\( I_s \)):
\[ I_s = \eta \times \frac{V_p \times I_p}{V_s} \]

6. **Substituting Known Values**:
Now, substituting the known values into the equation:
\[ I_s = 0.8 \times \frac{2400 \times 5}{240} \]

7. **Calculating the Output Current**:
First, calculate \( \frac{2400 \times 5}{240} \):
\[ \frac{2400 \times 5}{240} = \frac{12000}{240} = 50 \]
Now, substituting this back into the equation for \( I_s \):
\[ I_s = 0.8 \times 50 = 40 \text{ A} \]

### Final Answer:
The output current (\( I_s \)) is **40 A**.

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