Electrician Red Seal · Question
A 120 V AC circuit feeds a heater rated at 1500 W. An electrician replaces the heater with a new one that also operates at 120 V AC but is rated for 2000 W. Assuming both heaters are purely resistive, what is the approximate percentage change in the resistance of the new heater compared to the old one?
Resistance R = V²/P. For the old heater, R_old = (120 V)² / 1500 W = 14400 / 1500 = 9.6 Ω. For the new heater, R_new = (120 V)² / 2000 W = 14400 / 2000 = 7.2 Ω.
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Question: A 120 V AC circuit feeds a heater rated at 1500 W. An electrician replaces the heater with a new one that also operates at 120 V AC but is rated for 2000 W. Assuming both heaters are purely resistive, what is the approximate percentage change in the resistance of the new heater compared to the old one?
Answer options: ✅ Resistance decreased by 25%.
- Resistance increased by 33%.
- Resistance decreased by 33%.
- Resistance increased by 25%.
Correct answer: Resistance decreased by 25%.
Explanation: Resistance R = V²/P. For the old heater, R_old = (120 V)² / 1500 W = 14400 / 1500 = 9.6 Ω. For the new heater, R_new = (120 V)² / 2000 W = 14400 / 2000 = 7.2 Ω. The percentage change is ((R_new - R_old) / R_old) * 100% = ((7.2 - 9.6) / 9.6) * 100% = (-2.4 / 9.6) * 100% = -25%. Thus, the resistance decreased by 25%.
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