Resonace Frequency of a Parallel Resonant Circuit

For the network of Fig. 1, $$ Y_T = { 1 \over Z_1} + { 1 \over Z_2} + { 1 \over Z_4}\\ = { 1 \over R} + { 1 \over j X_{Lp}} + { 1 \over - j X_{C}}\\ ={ 1 \over R} -j { 1 \over X_{Lp}} + j{ 1 \over X_{C}}\\ Y_T={ 1 \over R} + j ({ 1 \over X_{C}} - { 1 \over X_{Lp}})\\ For unity power factor, the reactive component must be zero as defined by $${ 1 \over X_{C}} - { 1 \over X_{Lp}} = 0$$ Therefore, $${ 1 \over X_{C}} = { 1 \over X_{Lp}}$$ and $$X_{C} = X_{Lp}$$ Substituting for %X_{Lp}$ yields $${ 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}$$ or $$X^2_L = { L \over C} - R^2_l$$ $$2 \pi f_p L = \sqrt{{ L \over C} - R^2_l}$$ and $$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 Rl). Recognize also that as the magnitude of Rl approaches zero, fp rapidly approaches fs.