Question

Three 50omm resistors are connected in delta across a 40 v three phase supply if one of the resistor is disconnected then pashes current will be

Answer 1

Answer:

(i) Star Connection [Fig.] Vph = 400/√3V P = √3VLILcosφ = √3 x 400 x 4 x 1/√3 = 1600W (ii) Delta Connection Fig. Vph = 400V; Rph = 100Ω Iph = 400/100 = 4A IL = 4 x √3A P = √3 x 400 x 4 x √3 x 1 = 4800W When one of the resistors is disconnected (i) Star Connection [Fig.] The circuit no longer remains a 3-phase circuit but consists of two 100 Ω resistors in series across a 400-V supply. Current in lines A and C is = 400/200 = 2 A Power absorbed in both = 400 × 2 = 800 W Hence, by disconnecting one resistor, the power consumption is reduced by half. (ii) Delta Connection [Fig. ] In this case, currents in A and C remain as usual 120° out of phase with each other. Current in each phase = 400/100 = 4 A Power consumption in both = 2 × 42 × 100 = 3200 W (or P = 2 × 4 × 400 = 3200 W) In this case, when one resistor is disconnected, the power consumption is reduced by one-third.Read more on Sarthaks.com - https://www.sarthaks.com/494342/three-100-non-inductive-resistances-are-connected-in-star-delta-across-400-50-phase-mains?show=494353#a494353