Resistance:

$\overline{){\mathbf{R}}{\mathbf{=}}\frac{\mathbf{\rho}\mathbf{L}}{\mathbf{A}}}$

R_{1} = ρL/A

R_{2} = ρL/2A = R_{1}/2

R_{3} = ρL/3A = R_{1}/3

R_{4} = ρL/4A = R_{1}/4

Four wires are made of the same highly resistive material, cut to the same length, and connected in series.

- Wire 1 has resistance R
_{1}and cross-sectional area A. - Wire 2 has resistance R
_{2}and cross-sectional area 2A. - Wire 3 has resistance R
_{3}and cross-sectional area 3A. - Wire 4 has resistance R
_{4}and cross-sectional area 4A.

A voltage V_{0} is applied across the series, as shown in the figure.

Find the voltage V_{2} across wire 2.

Give your answer in terms of V_{0}, the voltage of the battery.

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Based on our data, we think this problem is relevant for Professor Soldatenko's class at Los Angeles Pierce College.