Loaded transformer basics: when the secondary voltage is one-fourth of the primary voltage, how does the secondary current compare to the primary current (ideal transformer)?

Introduction / Context:
Transformers trade voltage for current while ideally conserving power (minus losses). Understanding the inverse current–voltage relationship underpins power supply design and impedance matching.

Given Data / Assumptions:

• Secondary voltage Vs is one-fourth of primary voltage Vp.
• Ideal transformer assumption (no losses).
• Loaded condition (non-zero secondary current).

Concept / Approach:
For an ideal transformer, Vp/Vs = Np/Ns and Ip/Is = Ns/Np. Combining, Vs/Vp = Ns/Np, so Is/Ip = Np/Ns = Vp/Vs. If Vs = Vp/4, then Is = 4 * Ip, i.e., secondary current is four times the primary current.

Step-by-Step Solution:

Verification / Alternative check:
Check ideal power: Pp ≈ Vp * Ip ≈ Vs * Is = (Vp/4) * (4Ip) = Vp * Ip, consistent with ideal conservation.

Why Other Options Are Wrong:

• One-fourth the primary current / equal: contradicts inverse proportion for ideal transformers.
• Combined statement: logically inconsistent; cannot be both one-fourth and equal.

Common Pitfalls:

• Mixing up turns and current ratios (current ratio is inverse of voltage ratio).
• Forgetting that losses slightly alter real values—here the question is ideal.

Final Answer:
four times the primary current