A 12.0 V battery drives 1.00 ohm, 6.00 ohm and 13.0 ohm in series, producing a current of 0.600 A. What power does the 6.00 ohm resistor dissipate?

1. Power in a resistor is P = IV, and substituting V = IR turns this into P = I squared times R. 2. Every element of a series string carries the same current, here 0.600 A. 3. I squared is (0.600 A)(0.600 A) = 0.360 A squared. 4. For the 6.00 ohm resistor, P = (0.360 A squared)(6.00 ohm) = 2.16 W. 5. As a check, the three resistors dissipate 0.360 W, 2.16 W and 4.68 W, adding to 7.20 W, which equals the battery output of (0.600 A)(12.0 V). 6. 0.360 W is the power in the 1.00 ohm resistor, picked up by using the wrong resistance in the same formula. 7. 4.68 W is the power in the 13.0 ohm resistor, the largest share because the current is common and the resistance is largest. 8. 24.0 W is what this resistor would dissipate with the full 12.0 V across it, which is the parallel case, not the series one. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.1 Resistors in Series and Parallel_