Power Systems The equation of area control error(ACE) is ACE = ΔPtie - b.Δf ACE = ΔPtie + b.Δf ACE = ΔPtie - 1/b.Δf ACE = ΔPtie +1/ b.Δf ACE = ΔPtie - b.Δf ACE = ΔPtie + b.Δf ACE = ΔPtie - 1/b.Δf ACE = ΔPtie +1/ b.Δf ANSWER DOWNLOAD EXAMIANS APP
Power Systems A power system network has n nodes, Z33 of its bus impedance matrix is j0.5 pu. The voltage at node 3 is 1.3∠ -10°. If capacitor is having reactance of -j3.5 pu is now added to the network between node 3 and reference node, the current drawn by the capacitor in pu is 0.725∠ 80° 0.371∠ -100° 0.325∠ -100° 0.433∠ 80° 0.725∠ 80° 0.371∠ -100° 0.325∠ -100° 0.433∠ 80° ANSWER DOWNLOAD EXAMIANS APP
Power Systems If the percentage reactance of an element is 20 % and the full load current is 50 Amp, the short circuit current will be 200 Amp. 300 Amp. 350 Amp. 250 Amp. 200 Amp. 300 Amp. 350 Amp. 250 Amp. ANSWER DOWNLOAD EXAMIANS APP
Power Systems If all the sequence voltages at the fault point in a power system are equal, then the fault point is a LLL LLG LG LL LLL LLG LG LL ANSWER DOWNLOAD EXAMIANS APP
Power Systems A loss less transmission line having surge impedance loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30 %. Then SIL of compensated transmission line will be 3257 MW. 1835 MW. 2725 MW. 2280 MW. 3257 MW. 1835 MW. 2725 MW. 2280 MW. ANSWER DOWNLOAD EXAMIANS APP
Power Systems A transmission line has a reactance of 1 Pu is operating at Vs = Vr = 1 Pu. The generator is connected at source end which is delivering 0.5 Pu of active power and the transmission line is compensated with a series capacitance of 0.5 Pu. Find the load angle with series capacitance compensation ? 35.5° 29° 14.5° 10.5° 35.5° 29° 14.5° 10.5° ANSWER DOWNLOAD EXAMIANS APP
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