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Two wires of the same length made of two materials of resistivity $ρ_{1}$ and $ρ_{2}$ are connected in series to a source of potential $V$ . The equivalent resistivity for the combination is

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ρ_{1}

ρ_{2}

\frac{2 (ρ_{1} ρ_{2})}{ρ_{1} + ρ_{2}}

\frac{ρ_{1} + ρ_{2}}{2 ρ_{1} ρ_{2}}

\frac{ρ_{1} + ρ_{2}}{2}

answer is B.

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Detailed Solution

When two wires of the same length made of two materials of resistivity

ρ_{1}

and

ρ_{2}

are connected in series to a source of potential

V

, then the equivalent resistivity for the combination is

\frac{2 (ρ_{1} ρ_{2})}{ρ_{1} + ρ_{2}}

.
Equivalent resistance in parallel combination is given by

\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \dots (i)

Also, from the formula of resistance, the resistances

R_{1}

and

R_{2}

can be given as

R_{1} = ρ_{1} \frac{l}{A} and R_{2} = ρ_{2} \frac{l}{A}

Substituting value in

(i)

, we get

R_{p} = \frac{ρ_{1} ρ_{2} {(\frac{l}{A})}^{2}}{(ρ_{1} + ρ_{2}) \frac{l}{A}} = ρ \frac{l}{2 A}

\Rightarrow ρ = \frac{2 (ρ_{1} ρ_{2})}{ρ_{1} + ρ_{2}}

Hence, the equivalent resistivity for the combination is

\frac{2 (ρ_{1} ρ_{2})}{ρ_{1} + ρ_{2}}

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