To solve this problem, we use the current transformation rule for transformers:

$\frac{I_p}{I_s} = \frac{N_s}{N_p}$

Where:

• $I_p$: Primary current = $20 \, \text{A}$
• $I_s$: Secondary current (to be calculated)
• $N_p$: Number of primary turns = $600$
• $N_s$: Number of secondary turns = $200$

Rearranging the formula to find $I_s$: $I_s = I_p \cdot \frac{N_p}{N_s}$

Substituting the values: $I_s = 20 \cdot \frac{600}{200}$ $I_s = 20 \cdot 3 = 60 \, \text{A}$

Thus, the correct answer is:

2. 60A at secondary

---

Would you like me to explain this in further detail or clarify any concept?

5 Related Questions:

1. What is the relation between voltage and turns ratio in transformers?
2. How is power conserved in an ideal transformer?
3. What would happen if the secondary current exceeds its rated value?
4. How is the efficiency of a transformer calculated?
5. What are the main losses in a transformer?

Tip:

The primary-to-secondary current ratio is inversely proportional to the turns ratio in a transformer!