HWCh28Solutions

HWCh28Solutions - Physics 133 Homework solutions Chapter 28...

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Unformatted text preview: Physics 133 Homework solutions Chapter 28 P28.1 (a) P = V ( 29 2 R becomes 20.0 W = 11.6 V ( 29 2 R so R = 6.73 . (b) V = IR so 11.6 V = I 6.73 ( 29 and I = 1.72 A = IR + Ir so 15.0 V = 11.6 V + 1.72 A ( 29 r r = 1.97 . FIG. P28.1 P28.6 (a) R p = 1 1 7.00 ( 29 + 1 10.0 ( 29 = 4.12 R s = R 1 + R 2 + R 3 = 4.00 + 4.12 + 9.00 = 17.1 (b) V = IR 34.0 V = I 17.1 ( 29 I = 1.99 A for 4.00 , 9.00 resistors. Applying V = IR , 1.99 A ( 29 4.12 ( 29 = 8.18 V 8.18 V = I 7.00 ( 29 so I = 1.17 A for 7.00 resistor 8.18 V = I 10.0 ( 29 so I = 0.818 A for 10.0 resistor. FIG. P28.6 P28.11 (a) Since all the current in the circuit must pass through the series 100 resistor,...
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This note was uploaded on 06/02/2008 for the course PHY 133 taught by Professor K during the Fall '08 term at Cal Poly Pomona.

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HWCh28Solutions - Physics 133 Homework solutions Chapter 28...

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