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The original diagram has a series circuit containing $R$ and $L$, in parallel with $C$; it is not three separate parallel components. The electrical admittance is thereforeFor $R>0$, voltage and current are in phase when the imaginary part of $Y$ vanishes. Excluding the zero-frequency solution, division by $\omega$ givesA nonzero lossy parallel coil resonance exists only if $L>R^{2}C$. With $R=10\Omega$, $L=0.150\mathrm{H}$ and $C=1.00\times 10^{-5}\mathrm{F}$,For zero electrical resistance, the ideal parallel inductor-capacitor cancellation occurs atThe fractional reduction relative to the zero-resistance coil isAt exactly $R=0$ and $\omega ={\omega }_0$, $Y=0$, so finite applied voltage draws zero net current in the ideal model. The quoted frequency is the limiting parallel-resonance value; the phase of an exactly zero current is undefined.

Let $C_T=C+C_1$ be the total parallel capacitance, since capacitors in a parallel circuit have additive capacitances. Put ${\omega }_{\ast }=2\pi (50\mathrm{Hz})$ into the lossy parallel coil resonance condition:The additional capacitance is thereforeThis is the added capacitance, rather than the new total, and the conversion from frequency to angular frequency is essential.