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(A) In circuit $(A)$, the two resistors are in series. The total resistance $R_A$ is given by:$R_A = R_0(1 - \alpha \Delta T) + R_0(1 + \alpha \Delta

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$I_A$ remains constant while $I_B$ increases

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(A) In circuit $(A)$, the two resistors are in series. The total resistance $R_A$ is given by:$R_A = R_0(1 - \alpha \Delta T) + R_0(1 + \alpha \Delta T) = 2R_0$.Since the total resistance $R_A$ is independent of temperature, the current $I_A = V / R_A = V / (2R_0)$ remains constant.In circuit $(B)$, the two resistors are in parallel. The equivalent resistance $R_B$ is given by:$1 / R_B = 1 / [R_0(1 - \alpha \Delta T)] + 1 / [R_0(1 + \alpha \Delta T)]$$1 / R_B = [1 + \alpha \Delta T + 1 - \alpha \Delta T] / [R_0(1 - \alpha^2 \Delta T^2)] = 2 / [R_0(1 - \alpha^2 \Delta T^2)]$$R_B = R_0(1 - \alpha^2 \Delta T^2) / 2$.As temperature increases, $\Delta T$ increases, so $\alpha^2 \Delta T^2$ increases, which means $(1 - \alpha^2 \Delta T^2)$ decreases. Therefore, $R_B$ decreases.Since $I_B = V / R_B$, as $R_B$ decreases, the current $I_B$ increases.Thus, $I_A$ remains constant while $I_B$ increases.

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