For a series $$LCR$$  circuit, the power loss at resonance is

Solution :
In series $$LCR$$  circuit at resonance,
$${\text{capacitive reactance}}\left( {{X_C}} \right) = {\text{inductive reactance}}\left( {{X_L}} \right)$$
$${\text{i}}{\text{.e}}{\text{.}}\,\,\frac{1}{{\omega C}} = \omega L$$
Total impedance of the circuit
$$\eqalign{ & Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \cr & = \sqrt {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,Z = R \cr} $$
So, power factor, $$\cos \phi = \frac{R}{Z} = \frac{R}{R} = 1$$
Thus, power loss at resonance is given by
$$\eqalign{ & P = {V_{rms}}\,{i_{rms}}\cos \phi \,\,\left[ {\cos \phi = 1} \right] \cr & = {V_{rms}}\,{i_{rms}} \times 1 \cr & = \left( {{i_{rms}}R} \right){i_{rms}} \cr & = {\left( {{i_{rms}}} \right)^2}R = {i^2}R \cr} $$