$A$ sinusoidal voltage of peak value $283 \;V$ and frequency $50 \;Hz$ is applied to a series $LCR$ circuit in which $R = 3 \;\Omega, L = 25.48 \;mH,$ and $C = 796 \;\mu F.$ Find
$(a)$ the impedance of the circuit;
$(b)$ the phase difference between the voltage across the source and the current;
$(c)$ the power dissipated in the circuit; and
$(d)$ the power factor.

(A) To find the impedance of the circuit,we first calculate $X_{L}$ and $X_{C}$.
$X_{L} = 2 \pi \nu L = 2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} \; \Omega = 8 \; \Omega$.
$X_{C} = \frac{1}{2 \pi \nu C} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}} = 4 \; \Omega$.
Therefore,$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{3^{2} + (8 - 4)^{2}} = 5 \; \Omega$.
$(b)$ Phase difference,$\phi = \tan^{-1} \left( \frac{X_{L} - X_{C}}{R} \right) = \tan^{-1} \left( \frac{8 - 4}{3} \right) = \tan^{-1} \left( \frac{4}{3} \right) \approx 53.1^{\circ}$.
Since $\phi$ is positive,the current lags the voltage.
$(c)$ The power dissipated in the circuit is $P = I_{rms}^{2} R$.
$I_{rms} = \frac{V_{m}}{\sqrt{2} Z} = \frac{283}{\sqrt{2} \times 5} \approx 40 \; A$.
Therefore,$P = (40)^{2} \times 3 = 4800 \; W$.
$(d)$ Power factor = $\cos \phi = \frac{R}{Z} = \frac{3}{5} = 0.6$.