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 IT RO So ΤΟ IR Is IT UL-380 V CARGA 1 P₁=950 W cos p1=0,5 ind. CARGA 2 P₂=1.140 W cos (2=1 CARGA 3 P3=760 W cos 3=0,5 cap.

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 IT RO So ΤΟ IR Is IT UL-380 V CARGA 1 P₁=950 W cos p1=0,5 ind. CARGA 2 P₂=1.140 W cos (2=1 CARGA 3 P3=760 W cos 3=0,5 cap.

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.61P

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