Two metallic wires of identical dimensions are connected is series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is:

(B) $\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$

(A) $\frac{\sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$

(C) $\frac{\sigma_{1}+\sigma_{2}}{2 \sigma_{1} \sigma_{2}}$

(D) $\frac{\sigma_{1}+\sigma_{2}}{\sigma_{1} \sigma_{2}}$

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Current Electricity

Two metallic wires of identical dimensions are connected is series. If

σ_{1}

and

σ_{2}

are the conductivities of the these wires respectively, the effective conductivity of the combination is:

Answer: (B)

For wires in series, resistances add:

R_{eq} = R_{1} + R_{2}

. For a wire of length

ℓ

, area

A

, and conductivity

σ

, resistance

R = \frac{ℓ}{σ A}

. The equivalent wire has length

2 ℓ

and area

A

, so

R_{eq} = \frac{2 ℓ}{σ _{eq} A}

. From

R_{eq} = R_{1} + R_{2}

\frac{2 ℓ}{σ _{eq} A} = \frac{ℓ}{σ _{1} A} + \frac{ℓ}{σ _{2} A}

. Simplifying:

\frac{2}{σ _{eq}} = \frac{1}{σ _{1}} + \frac{1}{σ _{2}} = \frac{σ _{1} + σ _{2}}{σ _{1} σ _{2}}

. Therefore,

σ_{eq} = \frac{2 σ _{1} σ _{2}}{σ _{1} + σ _{2}}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option