Class 12 NEET Physics Current Electricity Questions Part 2

Concept:In circuit A, resistors are in series.Their temperature coefficients cancel out, making total resistance constant.In circuit B, resistors are in parallel, and their combined resistance decreases with temperature.Explanation:For circuit A (series):The total resistance is $R_0(1+\alpha \Delta T) + R_0(1-\alpha \Delta T) = 2R_0$, which is independent of temperature.Therefore, the current $I_A = V/(2R_0)$ remains constant.For circuit B (parallel):The equivalent resistance is $\frac{R_0(1+\alpha \Delta T) \cdot R_0(1-\alpha \Delta T)}{R_0(1+\alpha \Delta T) + R_0(1-\alpha \Delta T)} = \frac{R_0(1-\alpha^2 (\Delta T)^2)}{2}$.As temperature rises, $\alpha^2 (\Delta T)^2$ increases, so the numerator decreases.Hence $R_{\text{eq}}$ decreases.Since $I_B = V/R_{\text{eq}}$, the current $I_B$ increases with temperature.Answer:$I_A$ remains constant while $I_B$ increases.So the correct option is B.