P28_007.PDF

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Chapter 28 - 28.7
+ ir =12V+(0 . 040 Ω)(50 A) = 14 V.
(b) P = i 2 r =(50A) 2 (0 . 040 Ω) = 100 W.
(c) P = iV = (50 A)(12 V) = 600 W .
(d) In this case V =
E
ir =12V
−
(0 . 040 Ω)(50 A) = 10 V and P = i 2 r = 100 W .
7. (a) The potential difference is V =
E−

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