Answer:
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
