Power Systems Condition for zero voltage regulation is cosφ = X/R tanφ = X/R cosφ = R/X tanφ = R/X cosφ = X/R tanφ = X/R cosφ = R/X tanφ = R/X ANSWER DOWNLOAD EXAMIANS APP
Power Systems An over current relay, having a current setting of 125% connected to a supply circuit through a current transformer of ratio 500/5. The plug setting multiplier for a fault current of 5 kA is 5 8 6 2 5 8 6 2 ANSWER DOWNLOAD EXAMIANS APP
Power Systems The positive sequence reactance will be equal to negative sequence if the equipment is generator. none of these. transformer and transmission lines. motor. generator. none of these. transformer and transmission lines. motor. ANSWER DOWNLOAD EXAMIANS APP
Power Systems Four identical alternators each are rated for 20 MVA, 11 KV having a subtransient reactance of 16% are working in parallel. The short circuit level at the busbar is 400 MVA 125 MVA 100 MVA 500 MVA 400 MVA 125 MVA 100 MVA 500 MVA ANSWER DOWNLOAD EXAMIANS APP
Power Systems The voltage and current instantaneous values are given as 5 sin (ωt + 30°) pu and 2 sin (ωt - 15°) pu respectively. Find the per unit active power? 3.535 pu 7 pu 5 pu 5.353 pu 3.535 pu 7 pu 5 pu 5.353 pu ANSWER DOWNLOAD EXAMIANS APP
Power Systems The positive, negative and zero sequence per unit impedance of two generators connected in parallel are X1 = 0.12 pu, X2 = 0.096 pu and X0 = 0.036 pu. For LG fault at generator terminals (with 1 pu voltage) the positive sequence current will be 10.936 pu 9.936 pu 7.936 pu 8.936 pu 10.936 pu 9.936 pu 7.936 pu 8.936 pu ANSWER DOWNLOAD EXAMIANS APP
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