Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 359: 116

Answer

The phase angle, $\theta$, of the circuit is $16.22^{\circ}$.

Work Step by Step

Since the total Impedance of the circuit, $Z$, is $110 + 32i$ ohms $(\Omega)$, the phase angle in degrees, $\theta$, will be $tan \theta = \frac{32}{110}$ $\theta$ = $\arctan$ $(\frac{32}{110})$ $\theta$ = $16.22^{\circ}$ The phase angle, $\theta$, of the circuit is $16.22^{\circ}$.
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