Ideal transformer behavior: if a power transformer’s secondary voltage is twice the applied primary voltage, how does the secondary current compare with the primary current (assume ideal conditions)?

Introduction / Context:
Transformers trade voltage for current while approximately conserving power in the ideal case. When the secondary voltage is raised relative to the primary, the available current on the secondary drops proportionally. This principle underlies distribution networks and power supplies where different voltage levels are required without changing power significantly (aside from losses).

Given Data / Assumptions:

• V_s = 2 * V_p (secondary voltage is twice primary voltage).
• Ideal transformer: P_in ≈ P_out (ignoring losses).
• Current relation follows the inverse of the voltage ratio.

Concept / Approach:
For an ideal transformer, V_s / V_p = N_s / N_p and I_s / I_p = N_p / N_s. If V_s is twice V_p, then N_s is twice N_p, so I_s must be half I_p to keep apparent power roughly the same (S ≈ V * I on each side). Thus stepping voltage up by 2 steps current down by 2.

Step-by-Step Solution:
Given: V_s / V_p = 2 → N_s / N_p = 2.Current ratio: I_s / I_p = N_p / N_s = 1 / 2.Therefore I_s = 0.5 * I_p.Select “One-half the primary current.”

Verification / Alternative check:
Check power balance: If primary has 100 V and 2 A (200 VA), secondary would have ~200 V and 1 A (200 VA). This validates the inverse relationship between voltage and current in the ideal model.

Why Other Options Are Wrong:
“Same” or “twice” contradicts the inverse proportionality. “No more than 10% less” is arbitrary and incorrect in the ideal case. “None” is invalid because a correct statement is provided.

Common Pitfalls:
Forgetting that real transformers have copper and core losses leading to small deviations; mixing up which side is stepped up; ignoring that phase and power factor also matter for AC power calculations.

Final Answer:
One-half the primary current