A worn appliance cord has a resistance of 2.00 ohm instead of the 0.100 ohm it should have. What power is dissipated in it when 10.0 A flows?

1. For supply wires the loss is P = I squared R_w, with R_w the resistance of the wires themselves. 2. Square the current: (10.0 A) squared = 100 A squared. 3. Multiply by the worn resistance: P = (100 A squared)(2.00 ohm) = 200 W. 4. A healthy cord at 0.100 ohm would dissipate only 10.0 W at the same current. 5. Twenty times the heat inside insulation designed for 10 W is far more than is safe. 6. 20.0 W is the trap of using P = I R rather than P = I squared R. 7. 400 W would follow from doubling the current before squaring it. _Source: OpenStax College Physics (CC BY 4.0), Ch 20 "Electric Current, Resistance, and Ohm's Law", section Thermal Hazards_