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Two conductors arc made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a holJow tube of outer diameter 2 mm and inner diameter lrrun. Find the ratio of resistanceR_(A)" to " R_(B). |
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Answer» SOLUTION :Resistance of A, `R_(A) =(rho l )/(A) = (rho l)/(PI r^(2))` `therefore R_(A)= (rho l )/(pi xx ((1 xx 10^(-3))/(2) )^(2) ) =(4 rho l )/(pi xx 10^(-6)) "" `... (1) Resistance of B, `R_(B) = (rho l )/(pi [ ((2 xx 10^(-3))/(2) ) - ((1 xx 10^(-3))/(2) )^(2) ] )` `= (4 rho l)/( pi [ 4 xx 10^(-6) - 1 xx 10^(-6) ] ) ` `therefore R_(B) = (4 rho l)/(pi xx 3 xx 10^(-6))` `rArr(R_(A))/(R_(B)) = (4 rho l)/(pi xx 10^(-6)) xx (pi xx 3 xx 10^(-6))/(4 rho l)` `therefore (R_(A))/(R_(B)) = (3)/(1)` `therefore R_(A) : R_(B) = 3 : 1 ` |
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