# RC Parallel Circuit

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In a parallel circuit, the sum of the current's phasors must match the total current phasor. This value is quite close to the phasor-based approximations. The equations and methods for solving a parallel RC circuit are quite similar to those used in RL parallel circuits.
RC parallel circuits may be resolved in much the same way as are parallel RL circuits. The [Fig. 1] below illustrates a parallel RC circuit.
Fig. 1: Parallel R-C network.
The figure above shows a composite diagram of the circuit conditions. The current phasors $I_R$ and $I_C$ are out of phase, therefore, phasor addition must be used to determine total current. The solving of a parallel RC circuit follows the method previously applied to parallel RL circuits. Refer to [Fig. 1]. 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
Voltage (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,
$$I - I_R - I_C = 0$$
$$I = I_R + I_C$$
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
$$E = I Z_T= (10A \angle 0^\circ) (1 Ω \angle -53.19^\circ\\ = 10 V \angle -53.19^\circ$$

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