Two wires of the same length and same area made of two materials of resistivity ρ1 and ρ2 are connected in series to a source of potential V. The equivalent resistivity for the combination is

A
$ρ_{1}$ + $ρ_{2}$
B
$\frac{ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}}$
C
$\frac{ρ_{1} + ρ_{2}}{ρ_{1} ρ_{2}}$
D
$\frac{ρ_{1} + ρ_{2}}{2}$

Solution:

When two wires of the same length and same area are 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

\frac{ρ_{1} + ρ_{2}}{2}

.
“It is known that the resistance of a wire is inversely proportional to the area of cross-section, and directly proportional to the length and resistivity if the material.” Therefore, mathematically it can be written as

R = R_{1} + R_{2}

\Rightarrow ρ \frac{2 l}{A} = ρ_{1} \frac{l}{A} + ρ_{2} \frac{l}{A}

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

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

Two wires of the same length and same area 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

Solution:

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