Current lagging voltage by more then 90 degrees
But what about the 180° gap in the middle?
This implies a negative real part of the impedance, i.e., a negative resistance.
Express the impedance $Z$ in polar form
$$Z = R + jX = |Z|e^{j\phi}$$
where
$$|Z| = \sqrt{R^2 + X^2}$$
and
$$\tan \phi = \frac{X}{R}$$
For $R > 0$
$$-90^{\circ} \lt \phi \lt 90^{\circ}$$
but for $R < 0$
$$90^{\circ} \lt \phi \lt 270^{\circ}$$
Note that a genuine negative resistance does not dissipate energy but, instead, is an energy source.
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Admin
Updated on November 26, 2020Comments
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Admin almost 3 years
According to Wikipedia https://en.wikipedia.org/wiki/Leading_and_lagging_current
An alternating current that reaches its maximum up to 90° behind the voltage producing it is said to be lagging.
and
An alternating current that reaches its maximum value up to 90° ahead of the voltage that producing it is said to be leading.
But what about the 180° gap in the middle? are we saying that e.g. current can neither be a phase difference of 97° behind the voltage? I know mathematically to find this phase we use an inverse tan function but why do we take the acute (positive or negative) angle and not an angle greater then +90° or less then -90°
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theo almost 9 yearsThat's correct. Power flow is effectively reversed, as in regenerative breaking, where a generator is used to provide a retarding torque whilst re-charging a battery.
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coder_86 almost 9 yearsYour two inequalities where do they come from? is it just by experience, since the maths does not indicate this, unless I am overlooking something?
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Alfred Centauri almost 9 years@Joseph, why do you think "the maths does not indicate this"?
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Alfred Centauri almost 9 years@Joseph, it just occurred to me that you're not familiar with angles measured in radians? I'll edit my answer.
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coder_86 almost 9 yearsThe maths does not indicate this since for any $\frac{X}{R}$ the value of $arctan(\frac{X}{R})$ can take a value between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ take for example the case when $R=-1$ and $X=1$ then for $\phi = \frac{-\pi}{4}$ we have $tan(-\frac{\pi}{4})=-1$ Obvolusly here $R<0$ but $\phi$ lies in your first range rather then your second. So we must be making the choice to let $\phi$ be in the second range, (in this case $\frac{3\pi}{4}$)
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Alfred Centauri almost 9 years@Joseph, I haven't used inverse tangent in my answer. When the real part of a complex number is negative, the complex number is located in the 2nd or 3rd quadrant of the complex plane from which it follows that $\frac{\pi}{2} \lt \phi \lt \frac{3\pi}{2}$. This is elementary.
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TZDZ almost 9 years@Joseph You probably know that the arctan isn't enough to calculate the angle of a complex number ?