The most straightforward way to analyze reactive networks is with the concept of complex impedance. Analogous to resistance, the complex impedance tells us the relationship between voltage and current - but being a complex number, it allows us to take account of the phase difference "for free".
The impedance of a capacitor is $Z = \frac{1}{j\omega C}$ (we use $j$ rather than $i$ for the square root of -1 to prevent confusion with current, $I$.). Similarly for an inductor, $Z = j\omega L$. And inductor with series resistance would have $ZZ = j\omega L + R$.
A parallel LC network has a complex impedance that can be written as
$$Z = \frac{1}{\frac{1}{j\omega C}+j\omega L+R} = \frac{j\omega C}{1 - \omega^2 LC +j\omega RC}$$
As you can see, the denominator goes to zero at resonance in the case of a perfect inductor; if there is a bit of series resistance in the inductor, it won't quite go to zero...
You can now plot this as a function of frequency $\omega$ to tell you how much current can flow through your circuit as a function of frequency. For an inductor of 100 uH, a capacitor of 100 uF, and a series resistor varying between 0.5 Ohm and 5 Ohm, I get the following plot:
Which I generated with this Matlab code:
omega=linspace(0,10e3*2*pi, 1000);
L = 100e-6;
C = 100e-6;
figure
for R = [0.5 5]
Z = 1j*omega*C./(1-omega.^2*L*C+1j*omega*R*C);
plot(omega/(2*pi), abs(Z))
hold on
end
title 'frequency response of LCR circuit'
xlabel 'frequency (Hz)'
ylabel '|Z| (Ohm)'
legend({'R=0.5', 'R=5'})
As you can see, the series resistor has a big impact on the answer.
And your question contains a paradox: "when the voltage is the same" implies that you are driving the filter with the resonant frequency - any other voltage being offered to the filter would not cause equal voltages on the two components...