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795_Physics ProblemsTechnical Physics

# 795_Physics ProblemsTechnical Physics - Chapter 28 P28.8...

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Chapter 28 135 P28.8 120 1 2 3 4 V eq = = + + + F H G I K J IR I A A A A ρ ρ ρ ρ A A A A , or I A A A A ρ A = + + + 120 1 1 1 1 1 2 3 4 V a f e j V I A A A A A A 2 2 2 1 1 1 1 120 29 5 1 2 3 4 = = + + + = ρ A V V a f e j . P28.9 If we turn the given diagram on its side, we find that it is the same as figure (a). The 20 0 . and 5 00 . resistors are in series, so the first reduction is shown in (b). In addition, since the 10 0 . , 5 00 . , and 25 0 . resistors are then in parallel, we can solve for their equivalent resistance as: R eq = + + = 1 2 94 1 10 0 1 5 00 1 25 0 . . . . c h . This is shown in figure (c), which in turn reduces to the circuit shown in figure (d). Next, we work backwards through the diagrams applying I V R = and V IR = alternately to every resistor, real and equivalent. The 12 94 . resistor is connected across 25.0 V, so the current through the battery in every diagram is I V R = = = 25 0 1 93 . . V 12.94 A . In figure (c), this 1.93 A goes through the 2 94 . equivalent resistor to give a potential difference of: V IR = = = 1 93 2 94 5 68 . . . A V a fa f . From figure (b), we see that this potential difference is the same across V ab , the 10 resistor, and the 5 00 . resistor. (b) Therefore, V ab = 5 68 . V . (a) Since the current through the 20 0
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