A series RL circuit contains two resistors and two inductors. The resistors are 27 Ω and 47 Ω . The inductors have inductive reactances of 50 Ω and 40 Ω . The applied voltage is 120 V. What is the voltage drop on the inductor that has 40 Ω of reactance?

A) 28.41 V
B) 41.2 V
C) 51.5 V
D) 67.81 V

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Spencer Haugen · Accepted Answer

Final Answer : B, Explanation :  The voltage drop across an inductor in a series RL circuit can be found using the formula V_L = (X_L / Z) * V_total, where X_L is the inductive reactance, Z is the total impedance, and V_total is the applied voltage. First, calculate the total impedance Z = √((R_total)^2 + (X_total)^2) = √((27 + 47)^2 + (50 + 40)^2) = √(74^2 + 90^2) = √(5476 + 8100) = √13576 ≈ 116.5 Ω. Then, the voltage drop across the inductor with 40 Ω reactance is V_L = (40 / 116.5) * 120 ≈ 41.2 V.

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A series RL circuit contains two resistors and two inductors. The resistors are 27 Ω and 47 Ω . The inductors have inductive reactances of 50 Ω and 40 Ω . The applied voltage is 120 V. What is the voltage drop on the inductor that has 40 Ω of reactance?A) 28.41 VB) 41.2 VC) 51.5 VD) 67.81 V

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