A series RL circuit contains two resistors and two inductors. The resistors are 33 Ω and 47 Ω . The inductors have inductive reactances of 60 Ω and 30 Ω . The applied voltage is 120 V. What is the voltage drop on the 33 Ω resistor?

A) 32.89 V
B) 46.84 V
C) 55 V
D) 60 V

To find the voltage drop across the 33 Ω resistor in a series RL circuit, first, calculate the total impedance (Z) of the circuit. The total resistance (R) is the sum of the individual resistances: R = 33 Ω + 47 Ω = 80 Ω. The total inductive reactance (X_L) is the sum of the individual inductive reactances: X_L = 60 Ω + 30 Ω = 90 Ω. The total impedance (Z) can be calculated using the formula Z = √(R^2 + X_L^2) = √(80^2 + 90^2) = √(6400 + 8100) = √14500 ≈ 120.42 Ω. The current (I) in the circuit can be found using Ohm's law: I = V / Z = 120 V / 120.42 Ω ≈ 0.996 A. The voltage drop (V_R) across the 33 Ω resistor can be calculated using Ohm's law: V_R = I * R = 0.996 A * 33 Ω ≈ 32.89 V.