RLC Parallel ac Network Configuration with example

Refer to Fig. 1. Phasor Notation As shown in Fig. 2. $Y_T$ and $Z_T$ Admittance diagram: As shown in Fig. 3. I
$$\begin{split} I &= { E \over Z_T} = E Y_T\\ &= (100 V \angle 53.13^\circ)(0.5 S \angle -53.13^\circ) \\ &= 50A \angle 0^\circ\\ \end{split}$$ $I_R$ , $I_L$ and $I_C$ $$\begin{split} I_R &= (E \angle \theta)( G \angle 0^\circ)\\ &=(100 V \angle 53.13^\circ)(0.3 S \angle 0^\circ) = 30 A \angle 53.13^\circ\\ I_L &= (E \angle \theta)( B_L \angle -90^\circ)\\ &=(100 V \angle 53.13^\circ)(0.7 S \angle -90^\circ)\\ &= 70 A \angle -36.87^\circ\\ I_C &= (E \angle \theta)( B_C \angle 90^\circ)\\ &=(100 V \angle 53.13^\circ)(0.3 S \angle 90^\circ)\\ &= 30 A \angle 143.13^\circ\\ \end{split}$$ Kirchhoff's current law: At node a, $$I - I_R - I_L - I_C = 0$$ or $$I = I_R + I_L+ I_C$$ Phasor diagram: The phasor diagram of Fig. 4 indicates that the impressed voltage E is in phase with the current $I_R$ through the resistor, leads the current $I_L$ through the inductor by $90^\circ$, and lags the current $I_C$ of the capacitor by $90^\circ$.
Power: The total power in watts delivered to the circuit is $$\begin{split} P_T &= EI \cos \theta_T= (100 V)(50 A) \cos 53.13^\circ \\ &= (5000 W)(0.6)= 3000 W\\ or \\ P_T &=E^2 G= (100 V)^2(0.3 S) = 3000 W \end{split}$$ Power factor: The power factor of the circuit is $$F_p = \cos \theta_T = \cos 53.13^\circ \\ = 0.6 \text{lagging}$$ Impedance approach: The current I can also be found by first finding the total impedance of the network: $$\begin{split} Z_T &= {Z_RZ_LZ_C \over Z_R + Z_L+ Z_C}\\ &= 2 Ω \angle 53.13^\circ\\ \end{split}$$ And then, using Ohm's law, we obtain $$I = {E \over Z_T} = {100V \angle 53.13^\circ \over 2 Ω \angle 53.13^\circ} = 50 A \angle 0^\circ$$