In a balanced Y-Y circuit shown in Figure 5, the magnitude of the phase voltage at the source is Vp = 2520 V (rms) and Van = 252020° V (rms). The load impedance is ZY = 135 + j55 N per phase and the wire impedance is Zw=85 + j65 per phase. All the voltages and currents are in rms in this problem. (a) Find the phasors for line currents IA, IB, Ic in rms. (b) Find the load voltage VAN = IAZY. Also find VBN, and VCN. (c) Find the line voltages VAB, VBC, VCA at the load. (d) Find the complex power SL of the load. ZW ZY Van a A IA

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In a balanced Y-Y circuit shown in Figure 5, the magnitude of the phase voltage at the source is \( V_P = 2520 \, \text{V (rms)} \) and \( V_{an} = 2520 \angle 0^\circ \, \text{V (rms)} \). The load impedance is \( Z_Y = 135 + j55 \, \Omega \) per phase, and the wire impedance is \( Z_W = 85 + j65 \) per phase. All the voltages and currents are in rms in this problem.

(a) Find the phasors for line currents \( I_A, I_B, I_C \) in rms.
(b) Find the load voltage \( V_{AN} = I_A Z_Y \). Also find \( V_{BN} \) and \( V_{CN} \).
(c) Find the line voltages \( V_{AB}, V_{BC}, V_{CA} \) at the load.
(d) Find the complex power \( S_L \) of the load.

**Figure 5 Description:**

The diagram illustrates a three-phase balanced Y-Y circuit. Here, three phases are shown:

- Phase A: Starts from node \( a \) with voltage source \( V_{an} \). It goes through wire impedance \( Z_W \) and ends at node \( A \) with load impedance \( Z_Y \).
- Phase B: Starts from node \( b \) with voltage source \( V_{bn} \). It goes through wire impedance \( Z_W \) and ends at node \( B \) with load impedance \( Z_Y \).
- Phase C: Starts from node \( c \) with voltage source \( V_{cn} \). It goes through wire impedance \( Z_W \) and ends at node \( C \) with load impedance \( Z_Y \).

Each phase has its respective line current \( I_A, I_B, \) and \( I_C \) flowing through the impedances.

Nodes \( n \) and \( N \) represent the neutral connections for the source and the load, respectively.
Transcribed Image Text:In a balanced Y-Y circuit shown in Figure 5, the magnitude of the phase voltage at the source is \( V_P = 2520 \, \text{V (rms)} \) and \( V_{an} = 2520 \angle 0^\circ \, \text{V (rms)} \). The load impedance is \( Z_Y = 135 + j55 \, \Omega \) per phase, and the wire impedance is \( Z_W = 85 + j65 \) per phase. All the voltages and currents are in rms in this problem. (a) Find the phasors for line currents \( I_A, I_B, I_C \) in rms. (b) Find the load voltage \( V_{AN} = I_A Z_Y \). Also find \( V_{BN} \) and \( V_{CN} \). (c) Find the line voltages \( V_{AB}, V_{BC}, V_{CA} \) at the load. (d) Find the complex power \( S_L \) of the load. **Figure 5 Description:** The diagram illustrates a three-phase balanced Y-Y circuit. Here, three phases are shown: - Phase A: Starts from node \( a \) with voltage source \( V_{an} \). It goes through wire impedance \( Z_W \) and ends at node \( A \) with load impedance \( Z_Y \). - Phase B: Starts from node \( b \) with voltage source \( V_{bn} \). It goes through wire impedance \( Z_W \) and ends at node \( B \) with load impedance \( Z_Y \). - Phase C: Starts from node \( c \) with voltage source \( V_{cn} \). It goes through wire impedance \( Z_W \) and ends at node \( C \) with load impedance \( Z_Y \). Each phase has its respective line current \( I_A, I_B, \) and \( I_C \) flowing through the impedances. Nodes \( n \) and \( N \) represent the neutral connections for the source and the load, respectively.
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