# RLC Parallel ac Network Configuration with example   Whatsapp  Refer to Fig. 1. Fig. 1: Parallel RLC network.
Phasor Notation As shown in Fig. 2. Fig. 2: Applying phasor notation to the network of Fig. 1.
$Y_T$ and $Z_T$
$$\begin{split} Y_T &= Y_R + Y_L+ Y_C\\ &= G \angle 0^\circ + B_L \angle -90^\circ+ B_C \angle 90^\circ\\ &= {1 \over 3.33Ω} \angle 0^\circ+ {1 \over 1.43Ω} \angle -90^\circ + {1 \over 3.33Ω }\angle 90^\circ\\ &= 0.3S \angle 0^\circ +0.7S \angle -90^\circ + 0.3S \angle 90^\circ\\ &= 0.3 S - j0.7 S + j0.3 S= 0.5 S \angle -53.13^\circ\\ Z_T &= { 1 \over Y_T} = { 1 \over 0.5S \angle -53.13^\circ} \\ &= 2Ω \angle 53.13^\circ \\ \end{split}$$
Admittance diagram: As shown in Fig. 3. Fig. 3: Admittance diagram for the parallel RLC network of Fig. 1
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 $90^\circ$. Fig. 4: Phasor diagram for the parallel RLC network of Fig. 1.
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 $$I = {E \over Z_T} = {100V \angle 53.13^\circ \over 2 Ω \angle 53.13^\circ} = 50 A \angle 0^\circ$$

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