RL Parallel ac Network Configuration with example

Refer to Fig. 1. Phasor Notation As shown in Fig. 2. $Y_T$ and $Z_T$ $$\begin{split} Y_T &= Y_R + Y_L\\ &= G \angle 0^\circ + B_L \angle -90^\circ\\ &= {1 \over 3.33Ω} \angle 0^\circ + {1 \over 2.5Ω} \angle -90^\circ\\ &= 0.3S \angle 0^\circ +0.4S \angle -90^\circ\\ &= 0.3 S - j0.4 S = 0.5S \angle -53.13^\circ\\ Z_T &= { 1 \over Y_T} = { 1 \over 0.5S \angle -53.13^\circ} \\ &= 2Ω \angle 53.13^\circ \\ \end{split}$$ Admittance diagram: As shown in Fig. 3. I
$$\begin{split} I &= { E \over Z_T} = E Y_T\\ &= (20 V \angle 53.13^\circ)(0.5 S \angle -53.13^\circ) \\ &= 10 A \angle 0^\circ\\ \end{split}$$ $I_R$ and $I_L$ $$\begin{split} I_R &= { E \angle \theta \over R \angle 0^\circ}= (E \angle \theta)( G \angle 0^\circ)\\ &=(20 V \angle 53.13^\circ)(0.3 S \angle 0^\circ) = 6 A \angle 53.13^\circ\\ I_L &= { E \angle \theta \over X_L \angle 90^\circ} = (E \angle \theta)( B_L \angle -90^\circ)\\ &=(20 V \angle 53.13^\circ)(0.4 S \angle -90^\circ)\\ &= 8 A \angle -36.87^\circ\\ \end{split}$$ Kirchhoff's current law: At node a, $$I - I_R - I_L = 0$$ or $$\begin{split} I &= I_R + I_L\\ 10 A \angle 0^\circ &= 6 A \angle 53.13^\circ + 8 A \angle 36.87^\circ\\ 10 A \angle 0^\circ &=10 A \angle 0^\circ \end{split}$$ Phasor diagram: The phasor diagram of Fig. 4 indicates that the applied voltage E is in phase with the current IR and leads the current $I_L$ by $90^\circ$.
Power: The total power in watts delivered to the circuit is $$\begin{split} P_T &= EI \cos \theta_T= (20 V)(10 A) \cos 53.13^\circ \\ &= (200 W)(0.6)= 120 W\\ or \\ P_T &= I^2R = {V_R^2 \over R}= V^2 G\\ &= (20 V)^2(0.3 S) = 120 W \end{split}$$ 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: $$\begin{split} Z_T &= {Z_RZ_L \over Z_R + Z_L}\\ &= {(3.33 \angle 0^\circ)(2.5 \angle 90^\circ) \over (3.33 \angle 0^\circ)+(2.5 \angle 90^\circ)}\\ &={8.9 \angle 90^\circ \over 4.164 \angle 36.87^\circ}\\ &= 2 Ω \angle 53.13^\circ\\ \end{split}$$ And then, using Ohm's law, we obtain $$I = {E \over Z_T} = {20V \angle 53.13^\circ \over 2 Ω \angle 53.13^\circ} = 10 A \angle 0^\circ$$