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- The 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 componentsHi, Help me with this circuit questionA 220 Ω resistor is in series with an 80 mH inductor. The AC source is 48 at 1 kHz. The power factor is ________. Hint: Draw the circuit and get Vs and Is
- A surge of 100 kV travels along an overhead line towards its junction with a cable. The surge impedance for the overhead line and cable are 400 ohms and 50 ohms respectively. The magnitude of the surge transmitted through the cable is ?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 KVARProblem jD.2 E, S0.2 j0.1 le J0.4 ee M.2 Eliminate nodes 3 and 4 from the circuit and find Ybus and show all work
- 5.5 kW, 1450 rpm. L-type equivalent circuit and parameters for a continuous sinusoidal phase of a star connected 3-phase ASM with line voltage 220 V and frequency 50 Hz are given as follows, ignoring core (iron) losses: R1=0.21 Ohm, R2=-0.18 Ohm, X1=X2-0.66 Ohm and Xm = 9.9 Ohm How many times the inrush current will be the rated current when the ASM is started directly? Answer options group 8.0815 2.6455 4.3763 7.1534 1.2558 6.3414 5.2280 3.7473Power factor correction is a method to do which of the following.Which one of the following statements is true: The purpose of the power-factor correction is to add inductive elements to keep real power at a minimum level. The purpose of the power-factor correction is to add resistive elements (typically resistances) to improve the total reactive power. The purpose of the power-factor correction is to add reactive components (typically capacitive) to establish a system power factor closer to unity. The purpose of the power-factor correction is to add resistive elements (typically resistances) to establish a system power factor closer to zero.
- The impedance Z1 = 6+j8, Z2 = 8 -j6 and Z3 = 10+j0 ohms measured at 50Hz from three branches of a parallel circuit. This circuit is fed from a 100-volt 50-Hz supply. A purely reactive (inductive or capacitive) circuit is added as the fourth parallel branch to the above three-branched parallel circuit so as to draw minimum current from the source. Determine the value of L or C to be used in the fourth branch and also find the minimum current.The steady-state voltage drop between the load and the sending end of the line seen in (Figure 1) is excessive Suppose that V-4950/0° V (rms). A capacitor is placed in parallel with the 192 kVA load and is adjusted until the steady-state voltage at the sending end of the line has the same magnitude as the voltage at the load end, that is, 4950 V (rms). The 192 kVA load is operating at a power factor of 0.8 lag. Part A Calculate the size of the capacitor in microfarads if the circuit is operating at 60 Hz. In selecting the capacitance, use the value that results in the lowest possible power loss in the line. Express your answer in microfarads to three significant figures. ▸ View Available Hint(s) Figure 1 of 1 202 1100 192 kVA 0.8 lag ΜΕ ΑΣΦ. 11 vec C= 22.8 Submit Previous Answers Incorrect; Try Again; 5 attempts remaining Provide Feedback ? FThe course EE463 Powerr system analysis Please solve with clear writing And with all steps without short answer