# Phase Lead and phase Lag

I find the concept of phase lag and and lead quite confusing.

Suppose we have two voltages

$$\V(\theta) = V_{max} \cdot \sin(\theta + 30^o)\$$

and

$$\V(\theta) = V_{max} \cdot \sin(\theta + 60^o)\$$

If frequency and maximum voltages are same

then according to the definition

"The leading alternating quantity is one which reaches it maximum or zero value earlier than the other quantity."

$$\V(\theta) = V_{max} \cdot \sin(\theta + 60^o)\$$ is leading the $$\V(\theta) = V_{max} \cdot \sin(\theta + 30^o)\$$ by 30 degree.

But what if we compare

$$\V(\theta) = V_{max} \cdot \sin(\theta + 89^o)\$$ and $$\V(\theta) = V_{max}\cdot \sin(\theta + 120^o)\$$ ?

Then which one is leading ?

For example if $$\ \theta\$$ is 1 degree then $$\V(\theta) = V_{max} \cdot \sin(1^o + 89^o) = V(\theta) = V_{max} \cdot \sin(90^o)\$$

and $$\V(\theta) = V_{max} \cdot \sin(1^o + 120^o)\$$ is $$\V(\theta) = V_{max} \cdot \sin(121^o)\$$

so it means that $$\V_{max} \cdot \sin(\theta + 89^o)\$$ is leading the $$\V_{max}\cdot \sin(\theta + 120^o)\$$ because $$\89^o\$$ reaches its maximum value earlier than $$\120^o\$$ ?

While on the other had $$\120^o\$$ is greater than $$\89^o\$$ in phase. So what is the case ? Which one of them is leading ?

The start time is arbitrary. What matters is the difference between the waveforms. $$\\theta + 120\$$ is 31 degrees ahead of $$\\theta + 89\$$ and always will be.

The second example in the question is considering a time when the +120 eave has just passed the point of interest, so the next to reach it is the +89 wave. But the +120 had already reached it in this example: it’s clearly leading.

Note that leading by more than a half cycle, I.e. 270 degrees which is more than 180 and geometrically equivalent to -90, is considered to be lagging by convention.