Two inductors are connected in parallel to a 120-V, 60-Hz supply. The values of the inductors are 1.5 H and 2.5 H. The total inductive reactance is _____.

A) 525.5 Ω
B) 533.49 Ω
C) 942.5 Ω
D) 1508 Ω

The total inductive reactance (X_total) of inductors in parallel can be found using the formula for parallel reactances: 1/X_total = 1/X_1 + 1/X_2, where X_1 and X_2 are the reactances of the individual inductors. The reactance of an inductor is given by X = 2πfL, where f is the frequency and L is the inductance. For the 1.5 H inductor, X_1 = 2π(60)(1.5) = 565.5 Ω. For the 2.5 H inductor, X_2 = 2π(60)(2.5) = 942.5 Ω. Using the parallel reactance formula: 1/X_total = 1/565.5 + 1/942.5, solving for X_total gives approximately 533.49 Ω.