The relation `(E_(s))/(E_(P)) = (n_(s))/(n_(P))` is applicable only be

Options

A. a.c. generator
B. d.c. generator
C. induction coil
D. stepup/down transformer

Answer

To solve the question regarding the relation \( \frac{E_s}{E_P} = \frac{n_s}{n_P} \), we need to understand the context in which this relation is applied.

### Step-by-Step Solution:

1. **Understanding the Terms**:
- \( E_s \) = Electromotive force (EMF) in the secondary coil.
- \( E_P \) = Electromotive force (EMF) in the primary coil.
- \( n_s \) = Number of turns in the secondary coil.
- \( n_P \) = Number of turns in the primary coil.

2. **Context of the Relation**:
- This relation is derived from Faraday's law of electromagnetic induction, which states that the induced EMF in a coil is proportional to the rate of change of magnetic flux through the coil.
- In transformers, this relation helps us understand how the voltage changes when the number of turns in the coils changes.

3. **Application in Transformers**:
- In a transformer, if we want to increase the voltage (step-up transformer), we need to have more turns in the secondary coil than in the primary coil, i.e., \( n_s > n_P \). This results in \( \frac{E_s}{E_P} > 1 \).
- Conversely, if we want to decrease the voltage (step-down transformer), we need to have fewer turns in the secondary coil than in the primary coil, i.e., \( n_s < n_P \). This results in \( \frac{E_s}{E_P} < 1 \).

4. **Conclusion**:
- Therefore, the relation \( \frac{E_s}{E_P} = \frac{n_s}{n_P} \) is applicable specifically to transformers, which are devices used to either step up or step down AC voltages.

### Final Answer:
The relation \( \frac{E_s}{E_P} = \frac{n_s}{n_P} \) is applicable only to transformers.

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