The question asks to determine the secondary current and the type of transformer. Let's solve it step by step:

Given Data:

• Primary voltage ($V_p$) = 220 V
• Secondary voltage ($V_s$) = 50 V
• Primary current ($I_p$) = 2 A
• Efficiency ($\eta$) = 75% = 0.75

Formula for efficiency:

$\eta = \frac{P_s}{P_p}$ where:

• $P_s$ = power in the secondary circuit = $V_s \cdot I_s$
• $P_p$ = power in the primary circuit = $V_p \cdot I_p$

Rewriting the efficiency formula to calculate $I_s$: $I_s = \frac{\eta \cdot V_p \cdot I_p}{V_s}$

Substituting the values:

$I_s = \frac{0.75 \cdot 220 \cdot 2}{50}$ $I_s = \frac{330}{50} = 6.6 \, \text{A}$

Conclusion for the type of transformer:

The primary voltage ($V_p$) is greater than the secondary voltage ($V_s$). Therefore, this is a step-down transformer.

Final Answer:

• Secondary current ($I_s$) = 6.6 A
• Type of transformer = Step-down transformer

Would you like further explanation or have additional questions?

5 Related Questions:

1. How is transformer efficiency calculated for different load conditions?
2. What is the difference between a step-up and a step-down transformer?
3. How do core losses affect transformer efficiency?
4. Can the power output of a transformer exceed its input? Why or why not?
5. What are the practical uses of step-down transformers in daily life?

Tip: Always double-check the efficiency values in transformer problems, as it directly affects the secondary current calculation.