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Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistances R_A to R_B |
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Answer» Solution :Here ` rho_1 = rho_2 =rho ` and ` l_1 = l_2 = l` Conductor A is a solid wire of DIAMETER D=1 mm, its cross-section area`A_1 = (PI D^2)/(4)` Conductor B is a hollow tube of OUTER diameter `D_1` = 2 mm and INNER diameter `D_2` = 1 mm, henceits cross-section area `A_2 = pi/4 (D_1^2 - D_2^2)` As `R= (rho l)/(A) `, hence `(R_A)/(R_B) = A_2/A_1 =(pi/3 (D_1^2 - D_2^2) )/(pi/4D^2) = (D_1^2 - D_2^2)/(D^2) = ( (2mm)^2 - (1mm)^2)/((1mm)^2) = 3/1` ` rArr R_A = 3R_B` |
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