Consider two circuits A and B, each with two resistors. One resistor has a positive temperature coefficient +α and the other -α. In circuit A the two are in series; in circuit B they are in parallel, across the same battery. At the initial temperature each resistance is R₀. As temperature increases, the correct statement about the currents I_A and I_B is:

Solution & Explanation

Let the two resistances at higher temperature be R₊ = R₀(1 + αΔT) and R₋ = R₀(1 − αΔT). Circuit A (series): R_eq = R₀(1 + αΔT) + R₀(1 − αΔT) = 2R₀. The +αΔT and −αΔT terms cancel exactly, so R_eq stays 2R₀ for all temperatures — the current I_A = V/2R₀ is constant. Circuit B (parallel): R_eq = (R₊·R₋)/(R₊ + R₋) = [R₀²(1 − α²ΔT²)]/(2R₀) = (R₀/2)(1 − α²ΔT²). As T rises, ΔT grows, so the factor (1 − α²ΔT²) decreases, lowering R_eq below R₀/2. Lower resistance means larger current, so I_B increases. Therefore I_A remains constant while I_B increases — answer (A). Key point: in series the linear ±αΔT terms cancel; in parallel a second-order (1 − α²ΔT²) term survives and shrinks the resistance.