795_Physics ProblemsTechnical Physics

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

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Chapter 28 135 P28.8 120 1234 V eq == + + + F H G I K J IR I AAAA ρρρρ , or I ρ A = +++ 120 1111 V af ej V I A A 2 2 2 120 29 5 = A V V . 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 500 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 = ++ = 1 294 1 10 0 1 500 1 25 0 .. . . ΩΩΩ ch . 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 VI R = 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 0 193 . . V 12.94 A. In figure (c), this 1.93 A goes through the 2 94 equivalent resistor to give a potential difference of: ∆Ω R = 568 . A V . From figure (b), we see that this potential difference is the same across V ab , the 10 resistor, and the 500 resistor. (b) Therefore, V ab = V . (a) Since the current through the 20 0
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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