Classez les fils selon un ordre croissant en fonction de (a) la résistance et (b) la différence de potentiel à leurs bornes, sachant qu'un même courant circule dans chaque fil. Les... Classez les fils selon un ordre croissant en fonction de (a) la résistance et (b) la différence de potentiel à leurs bornes, sachant qu'un même courant circule dans chaque fil. Les dimensions des fils sont les suivantes : Fil 1 : diamètre = d, longueur = l Fil 2 : diamètre = 2d, longueur = 2l Read more

Steps to Solve

1. Calculate the area of the wires

The area of wire 1 will be calculated using the following formula:

$$A_1 = \pi r^2 = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$$

The area of wire 2 is:

$$A_2 = \pi (\frac{2d}{2})^2 = \pi d^2$$

2. Calculate the resistance of the wires

The resistance of a wire is given by:

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

Where $\rho$ is the resistivity of the material, $l$ is the length, and $A$ is the cross-sectional area.

Considering the resistance of wire 1 will be:

$$R_1 = \rho \frac{l}{\frac{\pi d^2}{4}} = \frac{4 \rho l}{\pi d^2}$$

Considering the resistance of wire 2 will be:

$$R_2 = \rho \frac{2l}{\pi d^2} = \frac{2 \rho l}{\pi d^2}$$

3. Compare resistances

Divide $R_1$ by $R_2$ to compare

$$\frac{R_1}{R_2} = \frac{\frac{4 \rho l}{\pi d^2}}{\frac{2 \rho l}{\pi d^2}} = 2$$

Therefore, $R_1 = 2R_2$. This means that the resistance of wire 2 is less than the resistance of wire 1.

4. Calculate the potential difference

Using Ohm's Law, the potential difference $V$ is given by:

$$V = IR$$

Since the current $I$ is the same in both wires:

$V_1 = IR_1$ and $V_2 = IR_2$

5. Compare potential differences

Since $R_1 = 2R_2$:

$V_1 = 2V_2$. This shows that the potential difference across wire 2 is less than the potential difference across wire 1.

6. Final Answer

Therefore, the wires in ascending order of resistance and voltage are: wire 2, wire 1.