R B Y R 400V, 60Hz IR Ze=550° Figure 4: Star Unbalanced Load IR ZR Is ly Zx ZR = 30 < 0° 0 Fault Zy17 <0° 0
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- Problem- Basics of Chopper & Types Calculate the R.M.S value of the fundamental component of current in a single phase Half bridge having a resistive load of 109 and supplied by an input DC voltage of 50V?two generators supplying a load. Generator I has a no-load frequency of 62.5 Hz and a slope Sp1 of I MW/Hz. Generator 2 has a no-load frequency of 62.0 Hz and a slope sp2 of I MW/Hz. The two generators are supplying a real load totaling 2.5 MW at 0.8 PF lagging. (a) At what frequency is this system operating, and how much power is supplied by each of the two generators? (b) Suppose an additional I-MW load were attached to this power system. What would the new system frequency be, and how much power would Gl and G2 supply now? Generator 1 VT V2 Generator 2 VTí KVAR KVARPower factor correction is usually done by adding _________ to the load circuit.Select the correct response:a. switchb. capacitorc. inductive loads
- With a D.C. generator which of the following regulation is preferred ? None of these 100% 50% 1%A. 3-8 50kw, 4-pole, soHZ inducffon motor haß a coinding Cacy desigmed for delta connection. the coinding has 24 conducforx por Mot mranged in Go slots, the mo value * the line current is 48A. Aind the fumdomental of the mmf coane f phase -A when the cunnent is passing throyph its maximum value. what ig the speed and peak volue the resultant mmf/ pule. folne il sume otheroise &leip. Al ame incorreet dorit copy fom Qne Anscwor Another AnsawerFrom the circuit below, determine secondary winding voltage Vsec if the primary winding voltage Vpri = 120Vrms? Np:Ns 6:1 Vp 120 Vac 60 Hz out R = 200 ) O Ovrms O 720Vrms 120Vrms O 20Vrms wi m
- 4 Q₁/ A 3-phase, 11 kv, 1.2 MVA, S.G its star connected- armature winding has effective resistance of 9% from Syn. impedance and syn. reactance of 50 52. Find VR% at half load (U.P.f and 0.8 lagging P.f.) of Raz0109 Zi 2₁²= R + X5² Z₁ = (010923) + X² 1 50 Zs: v Ra=01092₁ €54.9The instantaneous power absorbed by the load in a single-phase ac circuit, for a general R LC load under sinusoidal-steady-state excitation. is (a) Nonzero constant (b) Zero (c) Containing double-frequency componentsDoes an open connection permit balanced three-phase operation? - (a) Yes (b) No
- A single-phase, 120V(rms),60Hz source supplies power to a series R-L circuit consisting of R=10 and L=40mH. (a) Determine the power factor of the circuit and state whether it is lagging or leading. (b) Determine the real and reactive power absorbed by the load. (c) Calculate the peak magnetic energy Wint stored in the inductor by using the expression Wint=L(Irms)2 and check whether the reactive power Q=Wint is satisfied. (Note: The instantaneous magnetic energy storage fluctuates between zero and the peak energy. This energy must be sent twice each cycle to the load from the source by means of reactive power flows.)e = 0 Q1: What does happen if we change two currents? a. Q2: What does happen if we have a three-phase four poles machine? Q3: What does happen if we have a two-phase two poles machine? e = 0 Q4: What does happen if we apply a capacitance in phase b and then use same voltage for both phase. Dr. Ali Ki. pw --a step down chopper fed from 220V dc, is connected to RL load with R=10 ohms and LL=150 mH chopper frequency 1250HZ and duty cycle is 0.5 calculate the maximum value of load current (a)20.587A(b)8.521A (c)25.415A (d)12.478A