= Current reduction interval for shunt reactive compensation
{title2=$0<\omega CV^2<2Q_L$}
The <current reduction interval for shunt reactive compensation> is the range of ideal capacitive corrections that draws less source RMS <electric current> than the uncorrected inductive load. It is wider than the range that avoids <reactive power overcompensation>.
Assume fixed sinusoidal RMS <electric potential difference> $V>0$, <active power> $P>0$, and inductive <reactive power> $Q_L>0$. Let $q=\omega CV^2\ge0$ be the magnitude of the added capacitive <reactive power>. Comparing squared source <electric currents> gives
$$
I(q)^2-I(0)^2
=\frac{P^2+(Q_L-q)^2-P^2-Q_L^2}{V^2}
=\frac{q(q-2Q_L)}{V^2}.
$$
Consequently
$$
\boxed{I(q)<I(0)\quad\Longleftrightarrow\quad 0<q<2Q_L,}
$$
or $0<C<2C_*$ with $C_*=Q_L/(\omega V^2)$. The model's capacitive <reactive power> follows the sinusoidal <complex electrical power> convention.<AC Power Flow in Linear Networks by James L. Kirtley Jr.>[Energy and Power, equations 61–65 and 71]{ref}
At $q=Q_L$, source <electric current> is smallest. Between $Q_L$ and $2Q_L$, the source has a <leading power factor> but still draws less <electric current> than before correction. At $q=2Q_L$, the residual <reactive power> is $-Q_L$: its magnitude matches the original value, restoring the original source-current magnitude. Beyond that point the corrected <electric current> is larger.
For an original example, choose $P=300\,\mathrm W$, $Q_L=400\,\mathrm{var}$ and $V=100\,\mathrm V$. Uncorrected <electric current> is $5\,\mathrm A$. Full cancellation $q=400\,\mathrm{var}$ gives $3\,\mathrm A$. An overcorrection $q=600\,\mathrm{var}$ leaves $-200\,\mathrm{var}$ and draws only $\sqrt{13}\approx3.606\,\mathrm A$. But $q=800\,\mathrm{var}$ restores $5\,\mathrm A$, and $q=1000\,\mathrm{var}$ increases it to $\sqrt{45}\approx6.708\,\mathrm A$.
Two choices $q$ and $2Q_L-q$ are symmetric around the optimum and give the same squared source <electric current>. This symmetry explains why <power factor> magnitude alone cannot identify the sign of the residual <reactive power>. If $Q_L\le0$, no positive $q$ reduces current under the same fixed-load assumptions.
With the same upstream <electrical resistance>, reducing source mean-square <electric current> also reduces that conductor's <Joule heating>. It does not erase the current carried by the local <capacitor> branch or assess its ratings. Harmonic interactions, nonideal components and changing load conditions need their own model; this interval is an exact statement about the stated one-frequency circuit.
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