# RC Parallel ac Network Configuration with example

Refer to Fig. 1.
Fig. 1: Parallel R-C network.
Phasor Notation As shown in Fig. 2.
Fig. 2: Applying phasor notation to the network of Fig. 1.
$Y_T$ and $Z_T$ $$\begin{split} Y_T &= Y_R + Y_L\\ &= G \angle 0^\circ + B_C \angle 90^\circ\\ &= {1 \over 1.67Ω} \angle 0^\circ + {1 \over 1.25Ω} \angle 90^\circ\\ &= 0.6S \angle 0^\circ +0.8S \angle 90^\circ\\ &= 0.6 S + j0.8 S = 1.0 S \angle 53.13^\circ\\ Z_T &= { 1 \over Y_T} = { 1 \over 1.0 S \angle 53.13^\circ} \\ &= 1 Ω \angle -53.13^\circ \\ \end{split}$$
Admittance diagram: As shown in Fig. 3.
Fig. 3: Admittance diagram for the parallel R-C network of Fig. 1
E
$$\begin{split} E &= I Z_T = { I \over Y_T} \\ &= {(10 A \angle 0^\circ) \over (1 S \angle 53.13^\circ)} \\ &= 10 V \angle -53.13^\circ\\ \end{split}$$ $I_R$ and $I_C$ $$\begin{split} I_R &= (E \angle \theta) (G \angle 0^\circ)\\ &=(20 V \angle -53.13^\circ)(0.6 S \angle 0^\circ) = 6 A \angle -53.13^\circ\\ I_C &= ( E \angle \theta) (B_C \angle 90^\circ)\\ &=(10 V \angle -53.13^\circ)(0.8 S \angle 90^\circ)\\ &= 8 A \angle 36.87^\circ\\ \end{split}$$ Kirchhoff's current law: At node a, $$I - I_R - I_C = 0$$ or $$I = I_R + I_C$$ which can also be verified (as for the RC network) through vector algebra.
Phasor diagram: The phasor diagram of Fig. 4 indicates that the applied voltage E is in phase with the current $I_R$ and lags the capacitive current $I_C$ by $90^\circ$.
Fig. 4: Phasor diagram for the parallel R-C network of Fig. 1.
Power: The total power in watts delivered to the circuit is $$\begin{split} P_T &= EI \cos \theta_T= (10 V)(10 A) \cos 53.13^\circ \\ &= (200 W)(0.6)= 120 W\\ or \\ P_T &= E^2 G\\ &= (10 V)^2(0.6 S) = 60 W \end{split}$$ Power factor: The power factor of the circuit is $$F_p = \cos \theta_T = \cos 53.13^\circ \\ = 0.6 \text{leading}$$ Impedance approach: The voltage E can also be found by first finding the total impedance of the network: $$\begin{split} Z_T &= {Z_RZ_C \over Z_R + Z_C}\\ &= {(1.67 \angle 0^\circ)(1.25 \angle -90^\circ) \over (1.67 \angle 0^\circ)+(1.25 \angle -90^\circ)}\\ &={2.09 \angle -90^\circ \over 2.09 \angle -36.87^\circ}\\ &= 1 Ω \angle -53.13^\circ\\ \end{split}$$ And then, using Ohm's law, we obtain $$E = I Z_T= (10A \angle 0^\circ) (1 Ω \angle -53.19^\circ\\ = 10 V \angle -53.19^\circ$$