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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 components2. Two transformes A and B of ratings 5uokVA and 25okVA. Supplying a load kVA of 750 ato.8Pf lagging. open circunt valtajes are 400V and 410v respectively. Trusfarmn A has 1/ nsistance and 5/ ractance and traugformer B hay 1.$, mesistona and 41 reactance. Find (^) Crossacarent in the seandaries on no.load.ib) the load shared by euch transfarmer Their areThere is an installation with two equal triphasic loads of 300 kVA each and a power factor of 0.9 lagging. When connecting a pure reactive element in parallel, the reactive power at the entrance of the installation is 150 kVARS, with a lagging power factor. We can say that: a) A bank of reactors was connected and the fp deteriorated. b) A capacitor bank was connected and the pf deteriorated. c) A bank of reactors was connected and the fp improved. d) A capacitor bank was connected and the pf improved. e) None of the above
- 128, An equipment has an impedance 0.9 p.u. to a base of 20 MVA, 33 kV. To the base of 50 MVA, 11 kV, the p.u. impedance will be (a) 4.7 (c) 0.9 (b) 20.25 (d) 6.75In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0.. A load with active power= 100 MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 + j0.16 p.u on a base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 820) = 0•. Find |V2| () (one iteration). 100 MW 212 = 0.12 + j0.16 ++ 50 Mvar Vị = 1.0/0° О а. |V2| (1) = 0.4 p.u. Ob. |V2| (1) = 0.5 p.u. Oc. |V2| (1) = 0.7 p.u. d. (1) = 0.8 p.u. е. |V2| (1) = 0.6 p.u.The Full-looad and No-load of a regulator are 5.2V and 5.3V, respectively. Then the % load regulation is а. 0.1% b. 2.63% C. 1.92% d. 5.3%
- consider the following diagram of impedances in per unit of a common base a) get the YBUS admittance array and the ZBuz impedance array 213-j2 pu zt2-j0.15 zg2-j1.85 791-j0.9 211-0.1 1 212-j0.5pu 223-j0.5pu Va 1 p.u. Jv1-1p.u. 22-10 puj0.25 For the two coupled branches shown above, Ya =-j0.5, Yb = -j1 & Ym = j0.25. Using Equivalent Admittance, find the admittance located between nodes 2 & 3: O a. j0.25 O b. -j0.25 O c. None of these O d. -j1 O e. -j0.5An inductive circuit draws a line current of 10A. If the reactive component of the line current is 6A, then power factor of the circuit is ___ . a. 0.6 lagging b. 0.857 lagging c. 0.5 lagging d. 0.5 leading
- In an installation there is a three-phase load of 300 kVA with a power factor of 0.8 behind, in parallel with a three-phase load of 150 kVA with a power factor of 0.6 ahead. A pure reactive element is connected in parallel, and a unity power factor is obtained at the power input. Indicate the reactant of the element, with a positive sign if it is a bank of reactors, or with a negative sign if it is a bank of capacitors. Just put the corresponding sign followed by the numerical value, without units. The numerical value must correspond to the three-phase kVARS. You can use integer values with no decimal places or up to two decimal placesIn an installation there is a three-phase load of 300 kVA with a power factor of 0.8 behind, in parallel with a three-phase load of 150 kVA with a power factor of 0.6 ahead. A pure reactive element is connected in parallel, and a unity power factor is obtained at the power input. Indicate the reactant of the element, with a positive sign if it is a bank of reactors, or with a negative sign if it is a bank of capacitors. Just put the corresponding sign followed by the numerical value, without units. The numerical value must correspond to the three-phase kVARS. You can use integer values with no decimal places or up to two decimal placesHaving the below j5 -j10 -j6.8 j5 j0.8 j2.5 -j4.55 jo j1 j2.5 jo -j4.75] Nodal-Admittance Equation: j0.8 j1 j2.5 j2.5 Select one: a. jo b. j0.8 c. -j1.25 d. -j4.55 0 The impedance connecting node 3 to neutral is: Select one: O a. Infinity O b.-j0.8 O c. jo O d. 1.25 1.5625 L-53° 3.125 L36.87° The admittance connecting node 3 to neutral is: