Solve this in English please

For two resistors in parallel, $R_p = \frac{R_1 \times R_2}{R_1 + R_2}$.

The parallel combination is in series with the 3 $\Omega$ resistor.

The total resistance of the circuit is $\boxed{5 \, \Omega}$.

(iv) Calculate the current passing through X.

The voltage of the battery is $V = 12 \text{ V}$.
The total resistance of the circuit is $R_{total} = 5 \, \Omega$.

Ohm's Law states $V = I R_{total}$, so $I = \frac{V}{R_{total}}$.

The current passing through X (the total current) is $\boxed{2.4 \text{ A}}$.

9. (d) Resistance of a Wire

The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the cross-sectional area.

For the first wire, let its resistance be $R_1$, length $L_1 = L$, and cross-sectional area $A_1 = A$.
So, $R_1 = \rho \frac{L}{A}$.

For the second wire, it is made of the same material (so resistivity $\rho$ is the same), its length $L_2 = 3L$, and its cross-sectional area $A_2 = A$.
So, $R_2 = \rho \frac{3L}{A}$.

We can rewrite $R_2$ as:

Since $R_1 = \rho \frac{L}{A}$, we can substitute $R_1$ into the equation for $R_2$:

The second wire has three times the resistance of the first wire.