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A wire of given metarial having length l and ares of cross-section A has a resistance of `4 Omega`. What would be the resistance of another wire of the same meterial having length l/2 and area of cross-section 2A ? |
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Answer» For first wire `R_(1)=rhol/A=4Omega` Now for second wire `R_(2)= rho(l//2)/(2A)=1/4rhol/A` `R_(2)=1/4R_(1)` `R_(2)=1Omega` The resistance of the new wire is `1Omega`. |
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