Answer to Question #167017 in Electric Circuits for Richard

Question

An aluminum wire 7.5 m long is connected in parallel with a copper wire 6 m long. When a current of 5A is passed through the combination, it is found that the current in the aluminum wire is 3A.The diameter of the aluminum wire is 1.0 mm.Determine the diameter of the copper wire. Resistivity of copper is 0.0017 µΩ m; that of aluminum is 0.028 µΩm?

Answer 1

Resistance of the aluminum wire can be written as follows:

$$R_1=\rho_1\dfrac{L_1}{A_1}.$$

Resistance of the copper wire can be written as follows:

$$R_2=\rho_2\dfrac{L_2}{A_2}.$$

Let's divide $R_2$ by $R_1$, we get:

$$\dfrac{R_2}{R_1}=\dfrac{\rho_2}{\rho_1}\dfrac{L_2}{L_1}\dfrac{A_1}{A_2},$$ $$A_2=A_1\dfrac{R_1}{R_2}\dfrac{\rho_2}{\rho_1}\dfrac{L_2}{L_1}.$$

In the parallel circuit, the total current is the sum of the currents flowing through each component. Therefore, we can find the current flowing through the copper wire:

$$I_1+I_2=I_{tot},$$ $$I_2=I_{tot}-I_1=5\ A-3\ A=2\ A.$$

Also, in the parallel circuit, the voltage across each of the components is the same, therefore, we can write:

$$V=V_1=V_2,$$ $$I_1R_1=I_2R_2,$$ $$\dfrac{R_1}{R_2}=\dfrac{I_2}{I_1}=\dfrac{2}{3}.$$

Then, we can find the cross-sectional area of the copper wire:

$$A_2=\dfrac{\pi\cdot(1\cdot10^{-3}\ m)^2}{4}\cdot\dfrac{2}{3}\cdot\dfrac{0.017\ \mu \Omega m}{0.028\ \mu \Omega m}\cdot\dfrac{6\ m}{7.5\ m}=2.54\cdot10^{-7}\ m.$$

Finally, we can find the diameter of the copper wire:

$$A_2=\dfrac{\pi(d_2)^2}{4},$$ $$d_2=\sqrt{\dfrac{4A_2}{\pi}}=\sqrt{\dfrac{4\cdot2.54\cdot10^{-7}\ m}{\pi}}=5.69\cdot10^{-4}\ m=0.569\ mm.$$