7 m 3 r 4 m 4 r 243 10 4 m 0243 mm thus the diameter

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7 m 3 r 4 = 3.50 × 10 15 m 4 r = 2.43 × 10 4 m = 0.243 mm Thus, the diameter of the wire is 0.48 mm and the length is L = (3.366 × 10 7 m 1 )(2.43 × 10 4 m) 2 = 1.99 m Assess: It is reasonable to make a 1.99 m long wire with a diameter of 0.48 mm from an aluminum block of 1.0 g.
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32.39. Solve: The copper wire and the iron wire are connected in series. The composite resistance is simply the equivalent resistance of R Cu and R Fe : Cu Cu Fe Fe Cu Fe Cu Fe L L R R R A A ρ ρ = + = + Using the resistivity data from Table 31.2, ( ) ( ) ( ) ( ) ( ) ( ) 8 8 3 3 2 2 3 3 1.7 10 m 0.20 m 9.7 10 m 0.60 m 4.3 10 74 10 78 m 0.50 10 m 0.50 10 m R π π × Ω × Ω = + = × Ω + × Ω = Ω × ×
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32.40. Model: The wires and battery are ideal. Visualize: Solve: We can find the equivalent resistance necessary for the battery to deliver 9 W of power: ( ) ( ) ( ) 2 2 2 6.0 V 4.0 9.0 W V V P R R P Δ Δ = = = = Ω The combination of the 2.0 , 3.0 , and 6.0 Ω Ω Ω resistors that make 4.0 Ω is shown in the figure. The 3.0 and 6.0 Ω Ω parallel combination has an equivalent resistance of 2.0 , Ω which when added to the 2.0 Ω resistor in series totals 4.0 Ω equivalent resistance.
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32.41. Visualize: Solve: (a) The three resistors in parallel have an equivalent resistance of eq 1 1 1 1 12 12 12 R = + + Ω Ω Ω R eq = 4.0 Ω (b) One resistor in parallel with two series resistors has an equivalent resistance of eq 1 1 1 1 1 1 12 12 12 24 12 8 R = + = + = Ω + Ω Ω Ω Ω Ω R eq = 8.0 Ω (c) One resistor in series with two parallel resistors has an equivalent resistance of 1 eq 1 1 1 12 12 6 18.0 12 12 R = Ω + + = Ω + Ω = Ω Ω Ω (d) The three resistors in series have an equivalent resistance of 12 Ω + 12 Ω + 12 Ω = 36 Ω
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32.42. Model: Use the laws of series and parallel resistances. Visualize: Solve: Despite the diagonal orientation of the 12 Ω resistor, the 6 Ω , 12 Ω , and 4 Ω resistors are in parallel because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is 1 eq 1 1 1 2 6 12 4 R = + + = Ω Ω Ω Ω The trickiest issue is the 10 Ω resistor. It is in parallel with a wire , which is the same thing as a resistor with R = 0 Ω . The equivalent resistance of 10 Ω in parallel with 0 Ω is ( ) 1 1 eq 1 1 1 0 10 0 R = + = ∞ = = Ω Ω Ω In other words, the wire is a short circuit around the 10 Ω , so all the current goes through the wire rather than the resistor. The 10 Ω resistor contributes nothing to the circuit. So the total circuit is equivalent to a 2 Ω resistor in series with the 2 Ω equivalent resistance in series with the final 3 Ω resistor. The equivalent resistance of these three series resistors is R ab = 2 Ω + 2 Ω + 3 Ω = 7 Ω
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32.43. Model: Assume the batteries and the connecting wires are ideal. Visualize: Please refer to Figure P32.43. Solve: (a) The two batteries in this circuit are oriented to “oppose” each other. The curent will flow in the direction of the battery with the greater voltage. The direction of the current is counterclockwise because the 12 V battery is greater.
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