Answer to Question #287789 in Electrical Engineering for Joe

Question #287789

A one meter of annealed copper 2.5 cm in diameter is drawn until its resistance is 100 times the initial resistance, its diameter afterward is?


1
Expert's answer
2022-01-18T16:13:09-0500

The volume of this specimen is constant, the volume is the cross-sectional area A times length L:


V=A_1L_1=A_2L_2,\\\space\\ \dfrac{\pi d_1^2}{4}L_1=\dfrac{\pi d_2^2}{4}L_2→d_1^2L_1=d_2^2L_2,\\\space\\ [L_2]=L_1\dfrac{d_1^2}{d_2^2}.


The resistance:


R2=100R1, R=ρLA: 100ρL1A1=ρL2A2100L1A1=L2A2100L1d12=L2d22, d2=d110[L2]L1, d2=d110[L1d12d22]L1=d1210d2, d2=d110=0.79 cm.R_2=100R_1,\\\space\\ R=\dfrac {\rho L}A:\\\space\\ 100\dfrac {\rho L_1}{A_1}=\dfrac {\rho L_2}{A_2}→100\dfrac { L_1}{A_1}=\dfrac {L_2}{A_2}→100\dfrac {L_1}{d_1^2}=\dfrac {L_2}{d^2_2},\\\space\\ d_2=\dfrac{d_1}{10}\sqrt{\dfrac {[L_2]}{L_1}},\\\space\\ d_2=\dfrac{d_1}{10}\sqrt{\dfrac {\bigg[L_1\dfrac{d_1^2}{d_2^2}\bigg]}{L_1}}=\dfrac{d_1^2}{10d_2},\\\space\\ d_2=\dfrac {d_1}{\sqrt{10}}=0.79\text{ cm}.


The new diameter is 7.9 mm.


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