An ideal transformer has a 4:1 turns ratio (primary:secondary). If the measured input (primary) power is 5 W under load, what is the output (secondary) power delivered to the load (neglecting losses)?

Introduction / Context:For an ideal transformer, power in equals power out (P_in ≈ P_out), aside from small losses. While the turns ratio changes voltage and current levels, it does not create or destroy power. This question checks your understanding of power conservation across a transformer.

Given Data / Assumptions:

• Turns ratio Np:Ns = 4:1.
• Measured primary (input) power = 5 W.
• Neglect core and copper losses (ideal device).

Concept / Approach:Ideal transformer relations: V2/V1 = Ns/Np and I2/I1 = Np/Ns. Apparent power is conserved: P2 = V2 * I2 = V1 * I1 = P1. Therefore, the secondary power equals the primary power when losses are negligible.

Step-by-Step Solution:Given P_primary = 5 W.Assuming ideal behavior: P_secondary = P_primary.Thus, P_secondary = 5 W.

Verification / Alternative check:If the secondary voltage is V2 = V1 * (Ns/Np) = V1/4, then I2 = I1 * (Np/Ns) = 4 * I1. Multiplying gives V2 * I2 = (V1/4) * (4 * I1) = V1 * I1 = P1, confirming power equality independent of ratio.

Why Other Options Are Wrong:

• 1.25 W or 20 mW: Incorrect scaling; power is not reduced by the turns ratio in an ideal transformer.
• 9.5 W: Would imply power gain without an external source—violates energy conservation.
• 1.25 pW: Physically nonsensical given the stated input power.

Common Pitfalls:

• Mistaking voltage/current scaling for power scaling; turns ratio redistributes voltage and current while conserving power.
• Ignoring real-world losses; though small, they slightly reduce output below input in practice.

Final Answer:5.0 W