# Chapter 25 - Think and Solve: 44

a. 8.33 A. b. 83.3 V. c. 1.38 kW. d. Stepping up voltage reduces the fraction of power lost.

#### Work Step by Step

a. $$Electric \; power = (current) \times (voltage)$$ $$100,000 W = (I)(12000 V)$$ So I = 8.33 A. b. By Ohm's Law, $V = IR = (8.33 A)(10 \Omega) = 83.3 V$ voltage drop along each power line. c. In each line, $Electric \; power = (current) \times (voltage) = (8.3 A)(83.3 V) = 691 W$. In both lines, the total power wasted/dissipated as heat 2(691 W) = 1.38 kW. d. In this example, the 1.38 kW lost as heat is a few percent of the transmitted power. If the transmission voltage were 5 times lower, 2400 V, the current would be 5 times more, and the lost heat would be 25 times more! High-voltage power transmission is relatively efficient.

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