Given values:

Ratio of lengths of the wires: $\frac{L_1}{L_2} = \frac{2}{3}$

Ratio of radii of the wires: $\frac{r_1}{r_2} = \frac{8}{9}$

Since both wires are made of the same material and experience the same potential difference, we use Ohm's Law:

$ V = I R $

This leads to the relationship for the wires:

For the first wire: $V = I_1 R_1$

For the second wire: $V = I_2 R_2$

Equating the two since $V$ is the same:

$ I_1 R_1 = I_2 R_2 $

This implies:

$ \frac{I_1}{I_2} = \frac{R_2}{R_1} $

The resistance $R$ is given by:

$ R = \rho \frac{L}{A} $

where:

$\rho$ is the resistivity,

$L$ is the length,

$A = \pi r^2$ is the cross-sectional area.

Thus, for the resistances:

$ R_1 = \rho \frac{L_1}{\pi (r_1)^2}, \quad R_2 = \rho \frac{L_2}{\pi (r_2)^2} $

Substituting these into the current ratio equation:

$ \frac{I_1}{I_2} = \frac{\rho \frac{L_2}{\pi (r_2)^2}}{\rho \frac{L_1}{\pi (r_1)^2}} = \frac{L_2}{L_1} \times \left(\frac{r_1}{r_2}\right)^2 $

Substitute the given ratios:

$ \frac{I_1}{I_2} = \left(\frac{3}{2}\right) \times \left(\frac{8}{9}\right)^2 = \frac{3}{2} \times \frac{64}{81} = \frac{192}{162} = \frac{32}{27} $

Thus, the ratio of current $I_1 : I_2$ is $32 : 27$.

Current Electricity

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