3. For the A-A circuit below calculate phase and line currents: 0 A 173/120° V C b 173/0° V 173 /120° V 30 Ω j10 Ω 30 Ω j10 Ω 2000 m MUL C M 1100 30 Ω j10 Ω
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- For the single-phase circuit shown in Figure 122,I=100A. (a) Compute the phasors I1,I2, (b) Draw a phasor diagram showing I,I1,I2, and VThree loads are connected in parallel across a single-phase source voltage of 240V(RMS). Load 1 absorbs 15 kW and 6.667 kvar; Load 2 absorbs 3 kVA at O.96PF leading; Load 3 absorbs 15 kW at unity power factor. Calculate the equivalent impedance, Z, for the three parallel loads, for two cases: (i) Series combination of R and X, and (ii) parallel combination of R and X.Given V0 = 3.5 ∠122, V1 = 5.0 ∠ − 10, V2 = 1.9 ∠92, find the phase sequencecomponents VA, VB and VC.
- use only basic analysis techniques, series parallel wye delta I1, I2, I3, AND I4 are 4, 0.4, 2, 1.6 respectively.1. In a delta/delta system, with source phase- voltages of 100V and load phase resistances of 10Ω, calculate the apparent power in each phase. A. 1kVaB. 5kVaC. 50kVaD. 10kVa 2. Find the total true power for the circuit described in #1 A. 30kW B. 15kW C. 1.5kW D. 3kW1. In a delta/delta system, with source phase- voltages of 100V and load phase resistances of 10Ω, calculate the apparent power in each phase. A. 1kVaB. 5kVaC. 50kVaD. 10kVa 2. Find the total true power for the circuit described in #1 A. 30kWB. 15kWC. 1.5kWD. 3kW
- Draw the power system network and form bus admittance matrix (Ybus) matrix for the following power systems. (ii) Bus code Impedance Line charging admittance 1-2 4+j5 J0.05 1-3 6+j9 J0.02 2-3 13+j8 J0.03 2-4 6+j10 J0.05 3-4 12+j13 J0.06Electrical Circuits The sum of line currents must be zero. A 120V per phase, three-phase source delivers power to the following delta-connected load: - 1st Phase = 40∠0° - 2nd Phase = 20∠60° - 3rd Phase = 15∠45°. Determine the three line currents.For a three-phase positive sequence ΔΔ-connected system, Vab=3.6∠56.0Vab=3.6∠56.0. Then VbnVbn is:
- A 400V three-phase, four wire dystem has the following loads connected: Phase 1: Load impedance Z1= 4- + j15 Ohms Phase 2: Load impedance Z2 = 60 Ohms Phase 3: Load impedance Z3 = 15 - j20 Ohms Draw the circuit diagram and determine: The line and phase voltages for all phases (Including phase angles)The circuit shows an unbalanced electrical installation powered by a positive sequence symmetrical three-phase network of 380 V compound voltage. The loads are single phase. Load 1 absorbs an active power of 950 W with f.d.p. 0.5 inductive. Load 2 absorbs an active power of 1,140 W with f.d.p. Unit. Load 3 absorbs from the network an active power of 760 W with f.d.p. 0.5 capacitive. Taking the network voltage URN as the phase reference, calculate: a) complex expressions of the line currents IR. Is. and IT9. A single phase transmission line is delivering 500 kVA load at 2 kV. Its resistance is 0-2 Q and indpetivereactance is 0-4 2. Determine the voltage regulation if the load power factor is (i) 0-707 lagging (iü)0-707 leading.[(i) 5-3% () -165%]|