Example 12.12 Consider the unbalanced A-A circuit in Fig. 12.30. Use PSpice to find the generator current Lab, the line current IB, and the phase current Inc 12.9 PSpice for Three-Phase Circuits 531 a ww A 202 m 15 Ω 208/10° V € 502 2018/130 V b ww 202 m j5Q B -1402 208-110° V 130 Ω 202 j5Q ww -m C Figure 12.30 For Example 12.12.
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- The instantaneous power absorbed by the load in a single-phase ac circuit, for a general R LC load under sinusoidal-steady-state excitation. is (a) Nonzero constant (b) Zero (c) Containing double-frequency componentsSolve 8Q2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Ztoad 500 236.87° ohm , and the transmission line's impedance is Zine = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.
- 6.5% ans ex A so kvA. 2o00/20 V, So He single-phase transformar has impedance drof of8% and resistance drop of 4%. Caleulate. The Requlation of the Thams former at full- Load o8 P.s Legging. also find The Power factor at which Voltage requlation will Be Zero. Sols 0.5 leading. s alculate The % Noltege Requlation af atransformar in w hich the Percantage resistance drof is 1% and Percentage dre is 6% when The Power fador is 0.8 lagging ) > unity aMS 38 % ans 1 % (シ→ 0.8 leading . amS - 2.2%Q2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Zload = 500 236.87° ohm, and the transmission line's impedance is Zline = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.A single-phase power system is shown as the equivalent circuit below. The system has a 480V generator connected to an ideal 1:10 step-up transformer (no losses), a transmission line, an ideal 20:1 step-down transformer and a load all in series as shown. The resistance of the transmission line is 20 S2 and the reactance j60 , a load is connected to the system with an impedance of 10/30° №. For the analysis of the system, use base values of 10 kVA and 480 V at the generator, find (a) per unit values of all impedances, (b) per unit current, (c) load power (W), (d) power loss (W). NG 480 20⁰ V 10 kVA Zone 1 1:10 20 Ω j60 Ω ZL Zone 2 20:1 ILoad Zone 3 ZLoad 10/30° Ω
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- (b) A symmetrical delta-connected 415V three-phase supply is feed to an asymmetrical delta- connected load as shown in Figure Q2(b). Assume that it is a RYB sequence. The impedances of the load are ZRY = 50 +j50 Q , ZYB = 150 + j0 Q and ZBR= 50 – j50 N. (i) Determine and draw the phasor diagram of phase currents in the load. (ii) Determine and draw the phasor diagram of line currents in the system.A parallel connection of a RL branch with a C branch is connected across a 100V AC mains. At first R = 10 ohms, L = 20mH and frequency of 1000rad / s, the current measured is 2.2361A at 89.44 leading power factor. A) Determine the initial capacitance. B) However, a fault occurs on the capacitor branch making its capacitance 20% lower and a resistance of 5 ohms is detected. If this faulty circuit is rerun but at a frequency of 500 rad / s, determine the new current that will flow through the circuitA parallel connection of a RL branch with a C branch is connected across a 100V AC mains. At first R = 10 ohms, L = 20mH and frequency of 1000rad / s, the current measured is 2.2361A at 89.44 leading power factor. A) Determine the initial capacitance. B) However, a fault occurs on the capacitor branch making its capacitance 20% lower and a resistance of 5 ohms is detected. If this faulty circuit is rerun but at a frequency of 500 rad / s, determine the new current that will flow through the circuit.