Transformer loss accounting: With 120 W into the primary and 8.5 W lost to winding resistance, what output power is delivered to the load (neglect other losses)?

Introduction / Context:Practical transformers exhibit copper (I^2R) losses in windings and core losses. This question focuses on simple power balance when a single loss component is known, a common step in efficiency and thermal calculations.

Given Data / Assumptions:

• Primary input power Pin = 120 W.
• Winding copper loss Pcu = 8.5 W (given).
• Neglect all other losses (e.g., core loss), so remaining power is transferred to the load.

Concept / Approach:By conservation of power: Pout ≈ Pin - Losses. Here only one loss term is specified, so subtract it directly from the input power to estimate output power to the load terminals.

Step-by-Step Solution:

Verification / Alternative check:If we expressed efficiency η = Pout / Pin = 111.5 / 120 ≈ 92.9%, which is a plausible transformer efficiency figure depending on size and design.

Why Other Options Are Wrong:

• 14.1 W or 0 W: Ignore the stated power flow and losses.
• 1,020 W: Exceeds input power; violates conservation.
• 118.5 W: Would imply only 1.5 W loss, contradicting the given 8.5 W.

Common Pitfalls:

• Adding losses instead of subtracting from input power.
• Confusing efficiency percentage with watts.

Final Answer:111.5 W