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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 ?

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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.

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Try thinking about these:

motors tend to lag (when under load) ie the current is dragging it around

And

Generators tend to lead as the current is generated due to the movement of the armature...

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