A student sets up two circuits: Circuit X uses a thin, long copper wire, and Circuit Y uses a thick, short copper wire, both connected to identical batteries. Which circuit will have a higher resistance, and why?
Circuit X because the resistance increases with length and decreases with cross-sectional area
The resistance of a uniform metallic wire is given by \(R = \rho \dfrac{L}{A}\), where ρ is the resistivity of the material, L is the length and A is the area of cross-section.
From this relation, R is directly proportional to length (a longer wire has more collisions of electrons with lattice ions) and inversely proportional to the area of cross-section (a thicker wire gives electrons more parallel paths, easing the flow).
Circuit X uses a thin (small A) and long (large L) copper wire, so both factors act to increase R; Circuit Y uses a thick (large A) and short (small L) wire, giving it a much smaller resistance.
Hence, Circuit X has the higher resistance because resistance increases with length and decreases with cross-sectional area.
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