Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of $8\, ohms$, the resistance of the combination is equal to
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$\rho - {\rm{same, }}\,l - {\rm{same, }}\,{A_2} = \frac{1}{4}{A_1}$ (as ${r_2} = \frac{{{r_1}}}{2}$)
By using $R = \rho \frac{l}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{A_2}}}{{{A_1}}} \Rightarrow \frac{{{R_1}}}{8} = \frac{1}{4} \Rightarrow {R_1} = 2\,\Omega $
Hence, ${R_{eq}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{2 \times 8}}{{(2 + 8)}} = \frac{8}{5}\,\Omega .$
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