A gasoline-powered AC generator produces a maximum power of P = 5 kW. The peak v

Question

A gasoline-powered AC generator produces a maximum power of P = 5 kW. The peak voltage of its output is V = 120 V. You need to use this generator to drive a power tool that cannot take current larger than I_max = 6 A, so you build a transformer with N_p = 300 and N_s turns for the primary and secondary coils, respectively. What must be the minimum number of turns in the secondary coil N_s such that the maximum current I_max is not exceeded? a. N_s = 2100 b. N_s = 3400 c. N_s = 2900 d. N_s = 4800 e. N_s = 1600

Explanation / Answer

the current first = I = P/V = 5000/120 = 41.67 ampere

so, we know the relation:

NS = (300x41.67)/6.0 = 2083.5 = 2084 turns

so, the minimum number of turns should be 2100

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.