You want to use a metal bar as a resistor. Its dimensions are 1 by 3 by 13 units. To get the largest resistance from the bar, you should attach leads to the two opposite sides that have dimensions of 1 by 3 units. All configurations give the same resistance. 3 by 13 units. 1 by 13 units. None of the above.

1 by 3 units

Explanation:

The resistance (R) of a conductor is given by the formula:

R = ρL / A

where L is the length of the conductor, ρ is resistivity and A is the cross sectional area.

Let us assume that the metal bar has a resistivity of ρ.

a) If the leads is attached to the two opposite sides that have dimensions of 1 by 3 units.

The length of the bar would be 13 units and the cross sectional area (A) would be = 1 * 3 = 3 units²

R₁ = ρL / A =  ρ(13) / 3 = 13ρ / 3

b) If the leads is attached to the two opposite sides that have dimensions of 3 by 13 units.

The length of the bar would be 1 units and the cross sectional area (A) would be = 3 * 13 = 39 units²

R₂ = ρL / A  = ρ(1) / 39 = ρ / 39

c) If the leads is attached to the two opposite sides that have dimensions of 1 by 13 units.

The length of the bar would be 3 units and the cross sectional area (A) would be = 1 * 13 = 13 units²

R₃ = ρL / A = ρ(3) / 13 = 3ρ / 13

Therefore we can see that the largest resistance is gotten If the leads is attached to the two opposite sides that have dimensions of 1 by 3 units

area of focus