Two resistance wires are of same material having length, $l$ and $2l$ and area of cross-section, $4A$ and $A$ respectively. When they are connected in parallel, their equivalent resistance can be expressed as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{2}{9}(ρ\frac{l}{A})$
Resistance of a wire is given by
$R = \rho \frac{l}{A}$
For first wire:
$R_{1} = \rho \frac{l}{4A}$
For second wire:
$R_{2} = \rho \frac{2l}{A} = \frac{2\rho l}{A}$
When connected in parallel,
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{eq}} = \frac{4A}{\rho l} + \frac{A}{2\rho l}$
$\frac{1}{R_{eq}} = \frac{8A + A}{2\rho l}$
$\frac{1}{R_{eq}} = \frac{9A}{2\rho l}$
$R_{eq} = \frac{2\rho l}{9A}$