1. For each following load voltage and current pair, find the corresponding power factor and average power absorbed: a. v = 277√2 sin (377t + 30) V, i= 5.1 √2 sin (377t-10) A b. v = 679 sin (377t +50) V, i= 13 cos (377t-10) A c. v=-170 sin (377t-30) V, i= 8.1 cos (377t +30) A 2. What power is absorbed by a circuit that has an input admittance of 0.4 + j0.5 S and an input current of 30 A?

1. For each following load voltage and current pair, find the corresponding power factor and average power absorbed: a. v = 277√2 sin (377t + 30) V, i= 5.1 √2 sin (377t-10) A b. v = 679 sin (377t +50) V, i= 13 cos (377t-10) A c. v=-170 sin (377t-30) V, i= 8.1 cos (377t +30) A 2. What power is absorbed by a circuit that has an input admittance of 0.4 + j0.5 S and an input current of 30 A?

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.7MCQ

See similar textbooks

Similar questions

Let a 100V sinusoidal source be connected to a series combination of a 3 resistor, an 8 inductor, and a 4 capacitor. (a) Draw the circuit diagram. (b) Compute the series impedance. (C) Determine the current I delivered by the source. Is the current lagging or leading the source voltage? What is the power factor of this circuit?
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 components
Assume you are working on sinusoidal ac supply voltage which is in the form of 150 sin 220 t. So apply this voltage find out following parameters. Analyze the average value of voltage. Options: A)89.85volt B)90.75volt C)86.45volt D)95.55volt
Two three-phase alternators operate in parallel and the excitation is such that theinduced emfs are 1 kV/ph. The load admittance is (0,427 – j0,214) S. Theadmittance/phase of the machines are (0,0262 – j0,499) S and (0,0218 – j0,312) Srespectively. The emf of B leads A by 10°e. Calculate:(a) the terminal voltage(b) the current supplied by A(c) the power factor of B
Q2. A series of motor (Load 1) and lighting (load 2) are connected to an AC source as shown in Figure Q2. The AC source, Vs, supplies 415220° V with frequency 50 Hz. Based on this information: 5Ω 100mH : 15 2 Load 1 Load 2 i) Calculate the total load impedance. Vs ii) Calculate the current, I. Figure Q2. Write the waveform expressions in frequency domain for Vs and I. Draw a phasor diagram consists of Vs and / and state the relationship between Vs and I. (ii) (iv) Determine the power factor and calculate the real power, reactive power and apparent power. Establish the power triangle for the load.
Find the RMS value of Vit)-100 sin(377t+76) V 141,421 V 55,10 V 100 V 70.71 V
A 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 3 phase 50 Hz alternator is connected to the bus bars through an oil circuitbreaker. Due to short circuit occurring at bus bars, the circuit breaker opens when the r.m.s value of the current was 8000 amperes. The reactance of each phase of the alternator is 2.5 mH and the capacitance to earth between the circuit breaker and the alternator is 1.4 x 10^-7 uF. Determine: (i) Frequency of oscillations(ii) The maximum value of RRRV(iii) Time taken to achieve RRRV value in (ii)
15 A of current is taken by a non-inductive load at 125 V. In order that the same current shall be generated from 240 V, 50 Hz mains. An inductor is now also then connected in series. Note: The resistance of the inductor should be ignored and also assume the waveform to be sinusoidal. Determine the phase difference between the current and the applied voltage, the impedance of the circuit and the inductance of the inductor.
The total power drawn by the two-impedance load in parallel is 26kW at 0.98 power factor lagging, it is being supplied by a 230-volts AC source. One of the loads (Load 1) takes 16kVA at 0.77 power factor lagging. What is the reactive power of the unknown impedance (Load 2)?
Suppose I have two peak phasor voltages, V, = 100245° and V, = 75Z-60°. Circle Your Choices: I. The frequencies associated with the phasors are (0 Hz, 60 Hz, 50 Hz, Unknown). II. The rms value of the first phasor is (100, 70.7, 141, 200) volts. III. If you want to add these two phasors, the frequencies must be (60 Hz, 50 Hz, The same, Integer multiples of a given frequency.)
Prob. I and Vc For the following RMS phasor circuit, a coworker has determined values for 50 Ohms 50 Ohms i,- 225 2253.13° = 190.92 + 254.56 VRMS R1 + R2 Vc = 237.17/26.565° = 212.13+ 106.07 VRMS 100 2 -3Z98.13° =-0.4243 + J2.970 ARMS C -j60 Ohms For this problem provide at least 3 digits of precision for the answers. The results for the following calculations are to be answered in - a) Find the current I b) Determine the real (P) and reactive(Q) power DELIVERED by the source c) Determine the real (P) and reactive (Q) power ABSORBED by each impedance ( resistors R1 and R2, inductor and capacitor). The values for P and Q can be positive or negative. Remember S =P+jQ, so do not include the j with your answer for Q. Each of these answers will require two parts. For the phasor a magnitude and angle will be needed. For the powers, The real power P and the reactive power Q are required.