C in parallel is less than the resistance of e so the

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C in parallel is less than the resistance of E, so the total resistance along the path through A is less than the total resistance along path through D. The two paths have the same total potential difference—the emf of the battery—so more current will flow through the A path than through the D path. Consequently, A will have more current than D and E and will be brighter than D and E (A > D = E). Bulbs B and C each have half the current of A, because the current splits at the junction, so A is also brighter than B and C (A > B = C). The remaining issue is how B and C compare to D and E. Suppose B and C were replaced by wires with zero resistance, leaving just bulb A in the middle path. Then the resistance of the path through A would be half of the resistance of the path through D. This would mean that the current through A would be twice the current through D, so I A = 2 I D . When B and C are present, their resistance adds to the resistance of A to lower the current through the middle path. So in reality, I A < 2 I D . We already know that 1 B C A 2 I I I = = , so we can conclude that I B = I C < I D . Since the current through B and C is less than the current through D and E, D and E are brighter than B and C. The final result of our analysis is A > D = E > B = C.
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32.37. Visualize: Please refer to Figure P32.77. Solve: Bulb A is in parallel with the battery and experiences the full potential difference E whereas the other 5 bulbs divide up the potential into smaller pieces. So A will be brightest. Stated another way, the resistance of the right path, with bulb D in series with several other bulbs, is greater than the resistance of the middle path, with only bulb A. Both paths experience the full potential difference of the battery, so the current starting down the middle path is larger than the current starting down the right path, causing A to be brighter than D (A > D). All of the current in the right path passes through D, then it divides up. So D is brighter than B, C, E, or F (D > B, C, E, F). C and E are in parallel and have the same potential difference. Because they are identical bulbs with equal resistances, they have the same current and are equally bright (C = E). The current through F is the sum of the currents through C and E, so it is brighter than they are (F > C = E). The three-resistor combination C + E + F is in parallel with B. The combination C + E + F has more resistance than B, so more current will flow through B than through C + E + F. Consequently, B is brighter than F (B > F). Putting all these pieces together, the final result is A > D > B > F > C = E.
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32.38. Solve: The resistivity of aluminum is 2.8 × 10 8 Ω m and we want the wire to dissipate 7.5 W when connected to a 1.5 V battery. The resistance of the wire must be ( ) 2 2 2 1.5 V 0.30 7.5 W V V P R R P = = = = Ω Using the formula for the resistance of a wire, ( ) ( ) 8 7 1 2 2 0.30 2.8 10 m 3.366 10 m L L R L r A r ρ π = Ω = × Ω = × We need another relation connecting L and r . Making use of the mass density of aluminum, 3 3 2 7 3 2 1.0 10 kg 2700 kg/m 1.179 10 m r L r L π × = = × Using the value of L obtained above, r 2 (3.366 × 10 7 m 1 ) r 2 = 1.179 × 10
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