Two wires of the same dimensions but resistivities ${\rho _1}$ and ${\rho _2}$ are connected in series. The equivalent resistivity of the combination is
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(b) ${R_{eq}} = {R_1} + {R_2}$==>$\frac{{{\rho _{eff.2l}}}}{A} = \frac{{{\rho _1}l}}{A} + \frac{{{\rho _2}l}}{A}$$ \Rightarrow \,{\rho _{eff.}} = \frac{{{\rho _1} + {\rho _2}}}{2}$.
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