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)\$


\$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.


Try thinking about these:

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


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


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