Two metallic wires A and B are connected in parallel. Wire A has length l and radius r, wire B has a length 2l and radius 2r. Compute the ratio of the total resistance of parallel combination and the resistance of wire A.

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Solution

The resistance of a wire is directly proportional to its length and inversely proportional to the area of cross-section of the wire.
⇒ $R \propto \frac{l}{A} \propto \frac{l}{r^{2}}$
Wire A has length l and radius r and wire B has a length 2l and radius 2r. So, we can say,
$\frac{R_{A}}{R_{B}} = \frac{\frac{l}{r^{2}}}{\frac{2 l}{4 r^{2}}} = \frac{2}{1}$ .....(1)
Net resistance in parallel combination = $R_{r q} = \frac{R_{A} R_{B}}{R_{A} + R_{B}} = \frac{R_{A} \times \frac{R_{A}}{2}}{R_{A} + \frac{R_{A}}{2}} = \frac{1}{3}$ ...........(2)
The ratio of the total resistance of parallel combination and the resistance of wire A = $\frac{R_{e q}}{R_{A}} = \frac{1}{3}$ .

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