In an ideal transfomer, the voltage is step-down from 11 kV to 220 V. if the primary current be 100 A, the current in the secondary should be

Options

A. `1kA`
B. `0.5kA`
C. `2A`
D. `5kA`

Answer

To solve the problem, we will use the principle of conservation of power in an ideal transformer. The power in the primary side (input power) is equal to the power in the secondary side (output power).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Primary Voltage (\(V_p\)) = 11 kV = 11,000 V
- Secondary Voltage (\(V_s\)) = 220 V
- Primary Current (\(I_p\)) = 100 A

2. **Use the Power Conservation Principle:**
In an ideal transformer, the input power is equal to the output power:
\[ P_p = P_s \]
Where:
\[ P_p = V_p \times I_p \quad \text{(Power in primary)} \]
\[ P_s = V_s \times I_s \quad \text{(Power in secondary)} \]

3. **Set Up the Equation:**
Since \(P_p = P_s\), we can write:
\[ V_p \times I_p = V_s \times I_s \]

4. **Substitute Known Values:**
Substitute the known values into the equation:
\[ 11,000 \, \text{V} \times 100 \, \text{A} = 220 \, \text{V} \times I_s \]

5. **Calculate the Secondary Current (\(I_s\)):**
Rearranging the equation to solve for \(I_s\):
\[ I_s = \frac{V_p \times I_p}{V_s} \]
Substitute the values:
\[ I_s = \frac{11,000 \, \text{V} \times 100 \, \text{A}}{220 \, \text{V}} \]
\[ I_s = \frac{1,100,000}{220} \]
\[ I_s = 5000 \, \text{A} \]

6. **Conclusion:**
The current in the secondary (\(I_s\)) is 5000 A.

### Final Answer:
The current in the secondary should be **5000 A**.

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