Three loads are connected in parallel across a 1400-V rms, 60-Hz single-phase supply as shown in Figure 2.8. I 11 12 13 Load 1: Inductive load, 125 kVA at 0.28 power factor. Load 2: Capacitive load, 10 kW and 40 kvar. 1400 V 1 2 3 Load 3: Resistive load of 15 kW. (a) Find the total kW, kvar, kVA, and the supply power factor. (b) A capacitor of negligible resistance is connected in parallel with the above loads to improve the power factor to 0.8 lagging. Determine the kvar rating of this ca- pacitor and the capacitance in μF.

Three loads are connected in parallel across a 1400-V rms, 60-Hz single-phase supply as shown in Figure 2.8. Load 1: Inductive load, 125 kVA at 0.28 power factor. Load 2: Capacitive load, 10 kW and 40 kvar. Load 3: Resistive load of 15 kW. 1400 V (a) Find the total kW, kvar, kVA, and the supply power factor. 1 12 (b) A capacitor of negligible resistance is connected in parallel with the above loads to improve the power factor to 0.8 lagging. Determine the kvar rating of this ca- pacitor and the capacitance in uF. 13 3

In the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generators are connected in star with thier star points grounded. The system base is 15 MVA. The transformers all have reactances of 0.04 p.u. on this 15 MVA base. The reactances of all other elements are given in Table 1 (in 2) and the voltage levels are given in Table 2. p.u. G1 p.u. T1 jö Per-Unit Convert all values to p.u. on a 15 MVA base. Xa= p.u. Xc₂= XL = V BABE G1 2 X 9 T3 Figure 1: A section of the distribution system T1 L Table 1: Sequence reactances (2) 3 G1 L G2 0.3 0.59 0.01 4 L 9/10 10 Fault Voltage What is the voltage at bus 3 (in Volts) after the fault has occurred? Vp= V T2 5 T2 34 10/4 Table 2: Voltage bases (kV) G2 4 T3 10/9 | G2 Fault Current A three-phase fault with a fault reactance of 0.01 p.u. occurs at bus 3. Calculate the fault current flowing at the fault point in KA. Ip=-j KA So

Need asap pls help. Solve for the voltages across and currents through each component. Show the complete solution and explain. The total power should be zero..

What is the main direct cause of reactive power in AC system?A. Resistance of transmission linesB. Inductance and capacitance in the loadsC. Ideal transformer connected in the systemD. Power produced by generator

For the single-phase circuit shown in Figure 2.22, I5 10/08A. (a) Compute the phasors I1, I2, and V. (b) Draw a phasor diagram showing I, I1 I2, and V

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%

Q20. The section bus-bars A and B as shown in Figure B5 are linked by a bus-bar reactors each rated at 5000 kVA with 10% reactance. On bus-bar A, there are two generators each of 10,000 kVA with 10% reactance and on B two generators each of 8000 kVA with 12 % reactance. Find the Current and MVA that fed into a dead short circuit between all phases on B with bus-bar reactor in the circuit. Page 4 of 5 جامعة التقنية والعلومر التطبيقية بإبراء University of Technology and Applied Sciences - Ibra Student ID: Section: Student Name: 8,000 kVA 12% 10,000 kVA 10,000 kVA 10% 8,000 kVA 12% 10% 2 3 5,000 kVA 10% lell 8,000 kVA 10% Figure B5 "End of Question Paper"

Problem 1 - Series en Parallel AC networks [19] Look at the circuit in Figure 1 and determine the following: (a) Total Admittance. (b) Total Impedance. (c) Total Current (l:). (d) Current (I1) through impedance Z2. (e) Current (12) through impedance Z3. (f) Current (I3) through impedance Z4. (g) Is this an inductive or capacitive circuit? A. B Zs 220V;50HZ Figure 1 (h) Voltage across Z1. (i) Voltage across A and B. G) Voltage across Zs. Z1 = 3 + j5 ohm Z2 = 10 + jo ohm Z3 = 5 + j15 ohm Z4 = 10 – j30 ohm Zs = 20 – j30 ohm Admittance and Impedance in rectangular notation. All currents and voltage in polar notation. Take voltage as reference.

Driving-point admittance The actual impedance of a load is 30 OHM. The per-unit impedance of the load is 0.5 pu when Base Voltage is 600 V and Base Power is of O 6000 VA O 8000 VA O2000 VA 4000 VA Jump to ncements MacBook Air DD DII F10 F9 F8 000 F7 F6