8 Solution Set 8 on chapter 20

8 Solution Set 8 on chapter 20 - On Chapter 20 Electric...

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Unformatted text preview: On Chapter 20 Electric Circuits Problems pp. 634: # 2, 10, 21, 28, 39, 65, 71, 79, 84, 97, 104 Problems pp. 634: # 2, 10, 21, 28, 39, 65, 71, 79, 84, 97, 104 2. REASONING We are given the average current I and its duration Δ t . We will employ q I t ∆ = ∆ (Equation 20.1) to determine the amount Δ q of charge delivered to the ground by the lightning flash. SOLUTION Solving q I t ∆ = ∆ (Equation 20.1) for Δ q , we obtain ( 29 ( 29 3 1.26 10 A 0.138 s 174 C q I t ∆ = ∆ = × = ______________________________________________________________________________ 10. REASONING a. The resistance R of a piece of material is related to its length L and cross-sectional area A by Equation 20.3, / R L A ρ = , where r is the resistivity of the material. In order to rank the resistances, we need to evaluate L and A for each configuration in terms of L , the unit of length. Resistance Rank a 4 2 2 L R L L L ρ ρ = = × 1 b 1 2 4 8 L R L L L ρ ρ = = × 3 c 2 1 4 2 L R L L L ρ ρ = = × 2 1 Therefore, we expect that a has the largest resistance, followed by c , and then by b . b. Equation 20.2 states that the current I is equal to the voltage V divided by the resistance, / I V R = . Since the current is inversely proportional to the resistance, the largest current arises when the resistance is smallest, and vice versa. Thus, we expect that b has the largest current, followed by c , and then by a . SOLUTION a. The resistances can be found by using the results from the REASONING : a ( 29 2 2 2 2 1.50 10 m 0.600 5.00 10 m R L ρ-- = = × Ω ⋅ = Ω × b ( 29 2 2 1 1 1.50 10 m 0.0375 8 8 5.00 10 m R L ρ-- = = × Ω ⋅ = Ω × × c ( 29 2 2 1 1 1.50 10 m 0.150 2 2 5.00 10 m R L ρ-- = = × Ω ⋅ = Ω × × b. The current in each case is given by Equation 20.2, where the value of the resistance is obtained from part (a): a 3.00 V 5.00 A 0.600 V I R = = = Ω b 3.00 V 80.0 A 0.0375 V I R = = = Ω c 3.00 V 20.0 A 0.150 V I R = = = Ω ______________________________________________________________________________ 21. REASONING AND SOLUTION The mass m of the aluminum wire is equal to the density d of aluminum times the volume of the wire. The wire is cylindrical, so its volume is equal to the cross-sectional area A times the length L ; m = dAL ....
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8 Solution Set 8 on chapter 20 - On Chapter 20 Electric...

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