Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the

Question

Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the

Two metal wires of identical dimesnios are connected in series. If `sigma_(1)` and `sigma_(2)` are the conducties of the metal wires respectively, the effective conductivity of the combination is
A. `sigma_(1)+sigma_(2)`
B. `(sigma_(1)+sigma_(2))/(2)`
C. `sqrt(sigma_(1)+sigma_(2))`
D. `(2sigma_(1)+sigma_(2))/(sigma_(1)+sigma_(2))`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
Resistance of a wire in terms of conductivity `(sigma)` is given by `R=(1)/(sigma)(1)/(A)`
Where, l is the length and A in the area of cross-section of wire respectively.
As the wires are connected in series.
`therefore R_(s)=R_(1)+R_(2)Rightarrow (2l)/(sigma_(s)A)=(1)/(sigma(1)A)+(1)/(sigma_(2)A)`
Where, `simga`, is the effective conductivity
`(2)/(sigma_(s))=(1)/(sigma_(1))+(1)/(sigma_(2))=(sigma_(2)+sigma_(1))/(sigma_(1)sigma_(2))or sigma_(s)=(2sigma_(1)sigma_(2))/(sigma_(1)+sigma_(2))`