32CHAPTER 2. COMPLEX NUMBERSBABAphase shiftangleFigure 2.9: Phase shift between waves and vector phase anglecurrents in an AC circuit having deﬁnite phase angles. For example, if the current in an AC circuitis described as “24.3 milliamps at -64 degrees,” it means that the current waveform has an amplitudeof 24.3 mA, and it lags 64o behind the reference waveform, usually assumed to be the main sourcevoltage waveform.• REVIEW:• When used to describe an AC quantity, the length of a vector represents the amplitude of thewave while the angle of a vector represents the phase angle of the wave relative to some other(reference) waveform.2.3Simple vector additionRemember that vectors are mathematical objects just like numbers on a number line: they canbe added, subtracted, multiplied, and divided. Addition is perhaps the easiest vector operationto visualize, so we’ll begin with that. If vectors with common angles are added, their magnitudes(lengths) add up just like regular scalar quantities: (Figure 41,2.10)length = 6angle = 0 degreeslength = 8angle = 0 degreestotal length = 6 + 8 = 14angle = 0 degreesFigure 2.10: Vector magnitudes add like scalars for a common angle.Similarly, if AC voltage sources with the same phase angle are connected together in series, theirvoltages add just as you might expect with DC batteries: (Figure 42,2.11)Please note the (+) and (-) polarity marks next to the leads of the two AC sources. Even thoughwe know AC doesn’t have “polarity” in the same sense that DC does, these marks are essential toknowing how to reference the given phase angles of the voltages. This will become more apparentin the next example.If vectors directly opposing each other (180o out of phase) are added together, their magnitudes(lengths) subtract just like positive and negative scalar quantities subtract when added: (Figure 42,2.12)Similarly, if opposing AC voltage sources are connected in series, their voltages subtract as youmight expect with DC batteries connected in an opposing fashion: (Figure 42,2.13)Determining whether or not these voltage sources are opposing each other requires an exami-nation of their polarity markings and their phase angles. Notice how the polarity markings in the