Example 27.2 The Resistance of a Conductor Problem Calculate the resistance of a
ID: 776841 • Letter: E
Question
Example 27.2 The Resistance of a Conductor Problem Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectional area of 2.00 x 10-4 m2. Repeat the calculation for a cylinder of the same dimensions and made of glass having a resistivity of 3.0 x 1010 N·m. Solution From Equation 27.11 and Table 27.1, we can calculate the resistance of the aluminum cylinder as follows. 0.100 m R= P = (2.82 x 10-8 N . m) (2.00 x 10-4 m?) = 0.0000141 / 2 Similarly, for glass we find the following. ( 120 1010 , R= 0.100 m P = (3.0 x 1010 2 - m) (2.00 10-4 mi) 7 0.100 m ) = 1.5e13 As you might guess from the large difference in resistivities, the resistances of identically shaped cylinders of Jaluminum and glass differ widely. The resistance of the glass cylinder is 18 orders of magnitude greater than that| of the aluminum cylinder. Exercise 27.2 Hints: Getting Started | I'm Stuck What if the same amount of aluminum is used to make a cylinder with 2.0 times the original length. What will be the resistance of the resulting cylinder (in units of P2)? X u2 As the length increases, what else will change as a result, and how does these affect the resistance?Explanation / Answer
resistance R = rho*l/A*(l/l) = rho*l^2/(A*l) = rho*l^2/V
V = A*l = volume of given wire
as the same amount of wire is used
volume is same even if length is doubled
R1 = rh0*L1^2/V
R2 = rho*L2^2/V
R2/R1 = L2^2/L1^2
R2 = R1*(L2/L1)^2
given R1 = 0.0000141 ohm
L2 = 2*L1
R2 = 0.0000141*(2L1/L1)^2
R2 = 0.0000141*4
R2 = 56.4 micro ohm
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