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

Answer 1

Given:
E = 120 volts.
R1 = 27 Ohms.
R2 = 47 Ohms.
X1 = j50 Ohms.
X2 = j40 Ohms.
,
Z = (R1+R2) + j(X1+X2) = Total impedance.
Z = 74 + j90 = 116.5 Ohms[50.6o].

Z = Sqrt(74^2+90^2.
Tan A = 90/74.

I = E/Z = 120[0o]/116.5[50.6o] = 1.03Amps[-50.6o].
The negative angle means that the current lags the applied voltage by 50.6o.

V2 = I * X2 = 1.03[-50.6] * 40[90o] = 41.2 Volts & 39.4o.