In an ideal transformer, the voltagae and the current in the primary coil are 200V and 2 A, respectively. If the voltage in the secondary coil is 200 V, then the value of current in the secondary coil will be

Options

A. 0.2A
B. 2A
C. 10A
D. 20A

Explanation

To solve the problem, we will use the principle of conservation of energy in an ideal transformer, which states that the power in the primary coil is equal to the power in the secondary coil. This can be expressed with the formula:

\[ V_p \cdot I_p = V_s \cdot I_s \]

Where:
- \( V_p \) = Voltage in the primary coil
- \( I_p \) = Current in the primary coil
- \( V_s \) = Voltage in the secondary coil
- \( I_s \) = Current in the secondary coil

### Step-by-Step Solution:

1. **Identify the given values**:
- Voltage in the primary coil, \( V_p = 200 \, \text{V} \)
- Current in the primary coil, \( I_p = 2 \, \text{A} \)
- Voltage in the secondary coil, \( V_s = 200 \, \text{V} \)

2. **Write down the power conservation equation**:
\[ V_p \cdot I_p = V_s \cdot I_s \]

3. **Substitute the known values into the equation**:
\[ 200 \, \text{V} \cdot 2 \, \text{A} = 200 \, \text{V} \cdot I_s \]

4. **Calculate the left side**:
\[ 400 \, \text{VA} = 200 \, \text{V} \cdot I_s \]

5. **Solve for \( I_s \)**:
\[ I_s = \frac{400 \, \text{VA}}{200 \, \text{V}} = 2 \, \text{A} \]

6. **Conclusion**:
The current in the secondary coil, \( I_s \), is \( 2 \, \text{A} \).

### Final Answer:
The value of current in the secondary coil is **2 A**.

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