Question

An aluminium wire of 6.5metres long is connected in series with a copper wire 5 metre long. when a c voltage of 110 volts is connected accross the combination, it is found that the 65 volts is measured accros the aluminium wire. the diameter of the wite is 1mm. determine the diameter of the copper wire. resistivity of copper is 0.015 , that of aluminium is 0.023

Answer 1

The wires are connected in series, therefore their resistances add:

$$R_0=R_A+R_C=\frac{V_{110}}{I},\\ R_C=\frac{V_{110}}{I}-R_A.$$

On one hand, the current can be found from the voltage drop across the aluminum wire only because they are connected in series:

$$I=\frac{V_{65}}{R_A}=\frac{\pi d_A^2V_{65}}{4\rho_Al_A}.$$

On the other hand, the resistance of the copper and aluminum wires are respectively

$$R_C=\frac{4\rho_C l_C}{\pi d_C^2},\space R_A=\frac{4\rho_A l_A}{\pi d_A^2}.\\$$

Substitute this to the second equation above, do the same for the current (third equation):

$$\frac{4\rho_C l_C}{\pi d_C^2}=\frac{V_{110}}{I}-\frac{4\rho_A l_A}{\pi d_A^2},\\ \space\\ \frac{\pi d_A^2V_{65}}{4\rho_Al_A}\\ \space\\ \frac{4\rho_C l_C}{\pi d_C^2}=\frac{V_{110}}{V_{65}}\cdot\frac{4\rho_Al_A}{\pi d_A^2}-\frac{4\rho_A l_A}{\pi d_A^2}=\frac{4\rho_Al_A}{\pi d_A^2V_{65}}(V_{110}-V_{65}),\\ \space\\ d_C=d_A\sqrt{\frac{\rho_Cl_C}{\rho_Al_A}\cdot\frac{V_{65}}{V_{110}-V_{65}}}=0.85\text{ mm}.$$