Current through Capacitor and Inductor

Electrical Circuit Analysis > Resonance > Effect of Quality Factor greater than or Equal to 10

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$I_C$ 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 $X_C \approx X_L$ 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

$$ \bbox[10px,border:1px solid grey]{I_C \approx Q_l I_T}$$

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

$$ \bbox[10px,border:1px solid grey]{I_L \approx Q_l I_T}$$

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