Current through Capacitor and Inductor

I_C and I_L

A portion of Fig. 1 is reproduced in Fig. 2, with $I_T$ defined as shown.
Fig. 1
Fig. 2:Establishing the relationship between $I_C$ and $I_L$ and the current $I_T$.
As indicated, $Z_{T_{p}}$ at resonance is $Q_l^2 R_l$. The voltage across the parallel network is, therefore, $$V_C = V_L = V_R = I_T Z_{T_{p}} = I_T Q_l^2 R_l$$ The magnitude of the current $I_C$ can then be determined using Ohm's law, as follows: $$ I_C = {V_C \over X_C} = { I_T Q_l^2 R_l \over X_C} Substituting $X_C \appro X_L$ when $Q_l \geq 10$, $$ I_C = { I_T Q_l^2 R_l \over X_L} = I_T { Q_l^2 \over { X_L \over R_l }} = I_T { Q_l^2 \over Q_l}$$ and $$ \bbox[10px,border:1px solid grey]{I_C \appro Q_l I_T}$$ revealing that the capacitive current is $Q_l$ times the magnitude of the current entering the parallel resonant circuit. For large $Q_l$, the current $I_C$ can be significant.
A similar derivation results in $$ \bbox[10px,border:1px solid grey]{I_L \appro Q_l I_T}$$