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HWCh28Solutions

# HWCh28Solutions - Physics 133 Homework solutions Chapter 28...

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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    ε = 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,     P = I 2 R    P max = RI max 2 so    I max = P R = 25.0 W 100  = 0.500 A    R eq = 100  + 1 100 + 1 100 - 1   Ω= 150  V max = R eq I max = 75.0 V (b)    P = I V = 0.500 A (29 75.0 V (29 = 37.5 W  total power FIG. P28.11

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P 1 = 25.0 W P 2 = P 3 = RI 2 100  ( 29 0.250 A ( 29 2 = 6.25 W P28.16
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