How much reactive power does a load (30<85°) receive to which a current i(t) = 18 cos [400t + (-75%)] is applied? NOTE: Rounds to unity. Includes the corresponding units. Answer: Pa 93.77 var One possible correct answer is: 4842 VAR
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![How much reactive power does a load (30<85°) receive to which a current i(t) = 18 cos
[400t + (-75%)] is applied?
NOTE: Rounds to unity. Includes the corresponding units.
Answer:
Pa
93.77 var
One possible correct answer is: 4842 VAR](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbf7838de-33b1-4c9b-b444-de4b7c84d20a%2Fb4605ec3-c8df-4351-a54b-35f352816ddd%2Fxomli34_processed.jpeg&w=3840&q=75)
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- (b) Figure Q2(b) represents a single-line diagram of typical power system with the line impedances as indicated in pu on a 100 MVA base with a synchronous generator at Bus 1. The line charging susceptances are neglected. Calculate the bus voltages of V½ and V3 in pu using Gauss-Seidel method with the initial estimates of V, = 1+ j0 pu and V) = 1+ j0 pu. Perform only THREE (3) iterations. (0) (0) 1 230.5 0.01+j0.03 pu MW G 0.025+j0.05 pu 0.02+j0.04 pu 172.9 MVAR V1 = 1.0620°In the figure: R1 = 1439.170, R2 = R1 and R3 =2R2, then determine the value of the voltage source E. a 8 mA 10 mA 2 mA RT R1 R2 R3 O 25.33 11.51 O 17.27 15.43 13.82 +DC/AC MOTOR (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS LEGIBLY. NO LONG EXPLANATIONS NEEDED. ANSWER IN 2 DECIMAL PLACES. BOX THE FINAL ANSWER.) A current of (120-j150) A flows through a circuit when applied voltage is (8 + j12) V. A. What is the magnitude of the circuit impedance: a. 0.0635 ohm b. 0.0256 ohm c. 1.52 ohm d. 1.325 ohm B. What is the reactive power? a. 825.75 VAR b. 525.68 VAR c. 639.75 VAR d. 438.72 VAR
- O a72 O051 O 1.06 The value of armature current in the waveform is 14 A, a - 54 degrees, and B-155 degrees. The RMS value of current is la BeT O2.80 0497 O79.38 O10.49 直。Determine the rms current I as shown in figure 2. Source voltage is already in RMS value(Answer must be in polar formand should be in 4 decimal places).6. Determine the rms current I as shown in figure 1. Source voltage is already in RMs value (Answer must be in polar form and should be in 4 decimal places) -j4 n 20 j1.5 a B 40 200/30 V -12 n: Figure 1. 7. Determine the rms current I as shown in figure 2. Source voltage is already in RMS value (Answer must be in polar form and should be in 4 decimal places). j4n 240/50 V 火 js a 60 20 Figure 2.
- Please show you detailed solution Number 10 A voltage of 250 volts at 50 Hz is impressed on a circuit having a 60-ohm resistor, 35-?F capacitor, and a 0.25-henry inductor in series. Determine power factor. A. 0.979 C. 0.924 B. 0.862 D. 0.7076. Determine the rms current I as shown in figure 1. Source voltage is already in RMS value (Answer must be in polar form and should be in 4 decimal places). -j4 N j1.5 A B 4 0 A 200/30° V -j2 N: jIN Figure 1.PLEASE RESOLVE! THE FINAL ANSWER MUST BE IN THE CHOICES. DC/AC MOTOR (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS LEGIBLY. NO LONG EXPLANATIONS NEEDED. ANSWER IN 2 DECIMAL PLACES. BOX THE FINAL ANSWER.) A current of (120-j150) A flows through a circuit when applied voltage is (8 + j12) V. A. What is the magnitude of the circuit impedance: a. 0.0635 ohm b. 0.0256 ohm c. 1.52 ohm d. 1.325 ohm B. What is the reactive power? a. 825.75 VAR b. 525.68 VAR c. 639.75 VAR d. 438.72 VAR
- Find the average and rms value of the waveform shown in Figure-2. Em 3x ot- Figure-2QUESTION 11 An inductor is connected in parallel with a 6540 resistor and a function generator. The applied voltage is 42V RMS and the total current is 67mA RMS. Find the magnitude of the current through the inductor. Enter your answer in the space provided. Round your answer to one decimal place. Your answer should be in mA but do not enter the unit.3-9 An rms ac voltmeter and a peak-to-peak-reading ac voltmeter are to be used to determine whether three ac signals are sinusoidal. Deter. mine whether the signals are sinusoidal if the following readings are obtained First signal Second signal Third signal peak-to-peak reading = 35.26 V rms reading. - 12.00 V peak-to-peak reading - 11.31 V rms reading = 4.00 V peak-to-peak reading = 25.00 V rms reading -8.83 V
