An ideal transformer has `500` and `5000` turn in primary and secondary windings respectively. If the primary voltage is connected to a `6V` battery then the secondary voltage is

Options

A. `0`
B. `60V`
C. `0.6V`
D. `6.0V`

Explanation

To solve the problem, we need to understand the operation of an ideal transformer and the conditions under which it operates.

### Step-by-Step Solution:

1. **Identify the Turns Ratio**:
The transformer has a primary winding with \( N_p = 500 \) turns and a secondary winding with \( N_s = 5000 \) turns. The turns ratio \( \frac{N_s}{N_p} \) can be calculated as:
\[ \frac{N_s}{N_p} = \frac{5000}{500} = 10 \]

2. **Understand the Voltage Relationship**:
For an ideal transformer, the relationship between the primary voltage \( V_p \) and the secondary voltage \( V_s \) is given by:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Rearranging this gives:
\[ V_s = V_p \times \frac{N_s}{N_p} \]

3. **Determine the Primary Voltage**:
The primary voltage \( V_p \) is connected to a \( 6V \) battery, which means:
\[ V_p = 6V \]

4. **Calculate the Secondary Voltage**:
Substituting the values into the voltage relationship:
\[ V_s = 6V \times 10 = 60V \]

5. **Consider the Type of Current**:
However, it is crucial to note that transformers operate on alternating current (AC) and not direct current (DC). When a DC voltage is applied to the primary winding, the transformer does not function as intended because it requires a changing magnetic field to induce voltage in the secondary winding.

6. **Conclusion**:
Since the primary voltage is from a DC source, the transformer will not induce any voltage in the secondary winding. Thus, the secondary voltage \( V_s \) is:
\[ V_s = 0V \]

### Final Answer:
The secondary voltage is \( 0V \).