A transformer has an efficiency of 92.5%. the ratio number of turns in the primary coil to that in the…

Analyze the question: A non-ideal transformer has an efficiency of 92.5%. We know the turns ratio Np : Ns = 128 : 45 and the secondary current Is = 9.0 A. We need the primary current Ip.

Concept: Efficiency = (output power / input power) × 100%

Output power = Vs·Is and input power = Vp·Ip

For a transformer, Vs/Vp = Ns/Np, so Vs = Vp(Ns/Np).

Solution:

Efficiency = (Vs·Is) / (Vp·Ip)

0.925 = [Vp(Ns/Np)·Is] / (Vp·Ip)

0.925 = (Ns/Np)·(Is/Ip)

Make Ip the subject:

Ip = (Ns/Np)·Is / 0.925

Ip = (45/128) × 9.0 / 0.925

• 45/128 = 0.35156
• 0.35156 × 9.0 = 3.164
• 3.164 / 0.925 = 3.421

So Ip ≈ 3.42 A

Answer: Option C (3.42 A)

Note that 3.16 A (option B) is the ideal-transformer answer, where efficiency is ignored. Since the transformer loses some power, the primary must draw more current than that, so 3.42 A is correct.

Tip: For efficiency questions, always write the efficiency as output power ÷ input power, then substitute Vs/Vp = Ns/Np. The efficiency divides the ideal primary current, so the real primary current is always larger than the ideal one.