# Resonace Frequency of a Parallel Resonant Circuit

Whatsapp
For the network of Fig. 1,
Fig. 1:
$$\begin{split} Y_T &= { 1 \over Z_1} + { 1 \over Z_2} + { 1 \over Z_4}\\ &= { 1 \over R} + { 1 \over j X_{Lp}} + { 1 \over - j X_{C}}\\ &={ 1 \over R} -j { 1 \over X_{Lp}} + j{ 1 \over X_{C}}\\ Y_T&={ 1 \over R} + j ({ 1 \over X_{C}} - { 1 \over X_{Lp}})\\ \end{split}$$
For unity power factor, the reactive component must be zero as defined by
$${ 1 \over X_{C}} - { 1 \over X_{Lp}} = 0$$
Therefore,
$${ 1 \over X_{C}} = { 1 \over X_{Lp}}$$
and
$$X_{C} = X_{Lp}$$
Substituting for $X_{Lp}$ yields
$$\begin{split} { R^2_L + X^2_L \over X_L } &= X_C\\ R^2_L + X^2_L &= ( X_L )(X_C)\\ R^2_L + X^2_L &= ( wL ) {1 \over wC} = { L \over C} \end{split}$$
or
$$\begin{split} X^2_L &= { L \over C} - R^2_l\\ 2 \pi f_p L &= \sqrt{{ L \over C} - R^2_l} \end{split}$$
and
$$f_p = { 1 \over 2 \pi L } \sqrt{{ L \over C} - R^2_l}$$
Multiplying the top and bottom of the factor within the square-root sign by $C/L$ produces
$$f_p = { 1 \over 2 \pi L } \sqrt{{ 1 - R^2_l(C/L) \over C/L}}$$
$$f_p = { 1 \over 2 \pi L \sqrt{C/L}} \sqrt{1 - { R^2_l C \over L}}$$
and
$$\bbox[10px,border:1px solid grey]{f_p = { 1 \over 2 \pi \sqrt{LC}} \sqrt{1 - { R^2_l C \over L}}}$$
$$\bbox[10px,border:1px solid grey]{f_p = fs \sqrt{1 - { R^2_l C \over L}}}$$
where $fp$ is the resonant frequency of a parallel resonant circuit (for $Fp = 1$) and $fs$ is the resonant frequency as determined by $X_L = X_C$ for series resonance. Note that unlike a series resonant circuit, the resonant frequency $fp$ is a function of resistance (in this case $R_l$). Recognize also that as the magnitude of $R_l$ approaches zero, $fp$ rapidly approaches $fs$.

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