Question

When a resistor is connected across the terminals of an ac generator (112 V) that has a fixed frequency, there is a current of 0.500 A in the resistor. When an inductor is connected across the terminals of this same generator, there is a current of 0.400 A in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what are (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator?

For a)
I= E/R
0.500A= 112V/R -> R=224 ohms

0.400A= 112V/XL -> XL=280 ohms

Z= sqrt[R^2 + XL^2]
Z= 358.6 ohms
Did I do this correctly?

b) I'm not really sure about phasor diagrams. I know the current will lag the voltage, but I don't know how to determine the phase angle????

Answer 1

a. Correct.

b. tanA = Xl/R = 280/224 = 1.25
A = 51.3 Deg.

Since the circuit is inductive, the current lags the applied voltage by
51.3 Deg.