Question

Two wires of the same material having lengths in the ratio of $1:2$ and diameters in the ratio of $2:3$ are connected in series with an accumulator. Compute the ratio of the potential difference across the two wires.

Answer 1

Given: Two wires of the same material having lengths in the ratio of $1:2$ and diameters in the ratio of $2:3$ are connected in series with an accumulator.

To find the ratio of the potential difference across the two wires.

Solution:

Let $\rho$ be the resistivity of the material of the wire.

First wire:

Length = $L$

Diameter = $2d$

Area of Cross-section, A = $\pi {\left(\frac{2d}{2}\right)}^2=\pi d^2$

Resistance, $R=\rho \frac{L}{A}=\rho \frac{L}{\pi d^2}$

If current $I$ flows through the combination then the potential difference across R is, $V=IR$

Second wire:

Length = $2L$

Diameter = $3d$

Area of Cross-section, $A^{\prime }=\pi {\left(\frac{3d}{2}\right)}^2=\frac{9}{4}\times \pi d^2=\frac{9}{4}\times A$

Resistance, $R^{\prime }=\rho {\frac{2L}{A}}^{\prime }=\rho \frac{2L}{\frac{9}{4}A}=\rho \frac{8L}{9A}=\frac{8}{9}\rho \frac{L}{A}=\frac{8}{9}R$

If current $I$ flows through the combination then the potential difference across $R^{\prime }$ is, $V^{\prime }=I{R}^{\prime }=\frac{8}{9}IR$

So, the ratio of potential difference across the wires in series is,

$V:V^{\prime }=IR:\frac{8}{9}IR=9:8$