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A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IR. Assume the phase sequence to be RYB.(a) 44.74∠-63.4⁰A(b) 44.74∠63.4⁰A(c) 45.74∠-63.4⁰A(d) 45.74∠63.4⁰AThe question was posed to me in exam.I want to ask this question from Three-Phase Balanced Circuits in chapter Polyphase Circuits of Network Theory

Answer»

The correct OPTION is (a) 44.74∠-63.4⁰A

The explanation is: Taking the line VOLTAGE VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per PHASE = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IR = (400∠0^o)/(8.94∠63.4^o )= 44.74∠-63.4⁰A.



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