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Problem 6 Fun with Complex Power I 24020⁰ KV ~~ L3 Load, = 3000 kw, 4000 KVA Load₂= 1200 KVAR, Ⓒ = 30° Load3= = 2400 KVAR, pf =0,5 capacitive Find Total Average power IP, Total Reactive power Total Complex power S, Current Phasor I. Draw all 3 Load power triangles.

Problem 6 Fun with Complex Power I 24020⁰ KV ~~ L3 Load, = 3000 kw, 4000 KVA Load₂= 1200 KVAR, Ⓒ = 30° Load3= = 2400 KVAR, pf =0,5 capacitive Find Total Average power IP, Total Reactive power Total Complex power S, Current Phasor I. Draw all 3 Load power triangles.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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**Problem 6: Fun with Complex Power** Given the following electrical system: - Voltage Source: \(240 \angle 0^\circ\) kV - Loads: - Load 1: \(3000\) kW, \(4000\) kVA - Load 2: \(1200\) kVAR, \(\theta = 30^\circ\) - Load 3: \(2400\) kVAR, power factor = 0.5 capacitive **Tasks:** 1. Find the Total Average Power (\(P\)). 2. Find the Total Reactive Power (\(Q\)). 3. Calculate the Total Complex Power (\(S\)). 4. Determine the Current Phasor (\(I\)). 5. Draw all three Load Power Triangles. **Diagram Explanation:** The diagram includes three parallel loads \(L_1\), \(L_2\), and \(L_3\) connected to a voltage source of \(240 \angle 0^\circ\) kV. Each load has specific power specifications: - \(L_1\) is characterized by active and apparent power. - \(L_2\) is defined by reactive power and phase angle. - \(L_3\) has its reactive power defined and operates at a capacitive power factor. These loads are connected to a circuit with a current \(I\) flowing through them. The task involves calculating various power parameters and representing them graphically using power triangles for each load.
Transcribed Image Text:**Problem 6: Fun with Complex Power** Given the following electrical system: - Voltage Source: \(240 \angle 0^\circ\) kV - Loads: - Load 1: \(3000\) kW, \(4000\) kVA - Load 2: \(1200\) kVAR, \(\theta = 30^\circ\) - Load 3: \(2400\) kVAR, power factor = 0.5 capacitive **Tasks:** 1. Find the Total Average Power (\(P\)). 2. Find the Total Reactive Power (\(Q\)). 3. Calculate the Total Complex Power (\(S\)). 4. Determine the Current Phasor (\(I\)). 5. Draw all three Load Power Triangles. **Diagram Explanation:** The diagram includes three parallel loads \(L_1\), \(L_2\), and \(L_3\) connected to a voltage source of \(240 \angle 0^\circ\) kV. Each load has specific power specifications: - \(L_1\) is characterized by active and apparent power. - \(L_2\) is defined by reactive power and phase angle. - \(L_3\) has its reactive power defined and operates at a capacitive power factor. These loads are connected to a circuit with a current \(I\) flowing through them. The task involves calculating various power parameters and representing them graphically using power triangles for each load.
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