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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$. Since $L$ and $A$ are the same for all wires,the resistance is proportional to resist

Vedclass · Accepted Answer

$\frac{23\varepsilon}{12R_0}$

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$. Since $L$ and $A$ are the same for all wires,the resistance is proportional to resistivity $\rho$.Given that a wire with resistivity $\rho$ has resistance $R_0$,the resistances of the four wires are $R_1 = R_0$,$R_2 = 2R_0$,$R_3 = 4R_0$,and $R_4 = 6R_0$.These four wires are connected in parallel. The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$\frac{1}{R_{eq}} = \frac{1}{R_0} + \frac{1}{2R_0} + \frac{1}{4R_0} + \frac{1}{6R_0}$Taking the least common multiple $(LCM)$ of $1, 2, 4, 6$,which is $12$:$\frac{1}{R_{eq}} = \frac{12 + 6 + 3 + 2}{12R_0} = \frac{23}{12R_0}$Therefore,$R_{eq} = \frac{12R_0}{23}$.The total current $i$ is given by Ohm's law:$i = \frac{\varepsilon}{R_{eq}} = \frac{\varepsilon}{(12R_0 / 23)} = \frac{23\varepsilon}{12R_0}$.

If the resistivity of all four wires connected in parallel is different as shown in the figure,and all other dimensions of the wires are the same,then the total current $i$ is (the resistance of a wire having resistivity $\rho$ is $R_0$):

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