Question 26.65: A potential difference V is applied to a wire of cross-secti......
A potential difference V is applied to a wire of cross-sectional area A, length L, and resistivity ρ. You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 30.0 and the current is multiplied by 4.00. Assuming the wire’s density does not change, what are (a) the ratio of the new length to Land (b) the ratio of the new cross-sectional area to A?
We use P = i²R = i²ρL/A, or L/A = P/i²ρ.
(a) The new values of L and A satisfy
\left(\frac{L}{A}\right)_{\mathrm{new}}=\left(\frac{P}{l^{2}\rho}\right)_{\mathrm{new}}=\frac{30}{4^{2}}\left(\frac{P}{i^{2}\rho}\right)_{\mathrm{old}}=\frac{30}{16}\left(\frac{L}{A}\right)_{\mathrm{old}}.
Consequently, (L/A)_{\mathrm{new}}=1.875(L/A)_{\mathrm{old}}, and
{ L}_{\mathrm{new}}=\sqrt{1.875}L_{\mathrm{old}}=1.37L_{\mathrm{old}}~~\Rightarrow~~\frac{L_{\mathrm{new}}}{L_{\mathrm{old}}}=1.37\,.
(b) Similarly, we note that (L A)_{\mathrm{new}}=(L A)_{\mathrm{old}}, and
A_{\mathrm{new}}=\sqrt{1/1.875}\,A_{\mathrm{old}}=0.730\,A_{\mathrm{old}}\;\Rightarrow\;\frac{A_{\mathrm{new}}}{A_{\mathrm{old}}}=0.730\,.