Use the circuit reduction method to find the following quantities for the below circuit: a) Req- (answer: 174.1222) b) V. (11.89 V) c) V₁. (20.11 V) d) V₂. (11.89 V) e) I. (20.53 mA) Note that E 32 V is equivalent to E 32 V. Rea E 400 32 V 400 f www N 7100 Ω 600 220 2 220 f 1600 1220 In
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- Consider the load convention that is used for the RLC elements shown in Figure 2.2 of the text. A. If one says that an inductor absorbs zero real power and positive reactive power. is it (a) True (b) False B. If one says that a capacitor absorbs zero real power and negative reactive power (or delivers positive reactive power), is it (a) False (b) True C. If one says that a (positive-valued) resistor absorbs (positive) real power and zero reactive power, is it (a) True (b) False1 b b 9:1Y ۱۳ ۱ من ۱۳ ۱ For the circuit in figure below, he value of 12 when V2=0 V is 50 6 A sa 0.5 V 20 امجد جاسم کاظم/A/ 9:11p äclull 21.05.31 •.. :QUESTION 58 Short Problem: Given a parallel RLC circuit comprised of the following element: 94.42-ohm resistor, ideal inductor with reactance of 60.84 ohms; and a capacitor with reactance of 15.03 ohms. Compute for reactive power in VArs given an ac voltage source of 100 cis 0 volts. Give only the absolute value. Note: Follow this reminder carefully. Compute to the nearest 4 decimal places. No Scientific notation. Do not round off in the middle of calculation. Use stored values. Write the numerical values only. No units in your final answer. Spaces are not allowed. Excessive number of decimals as compared to the required number of decimals may result to an incorrect answer.
- Given a series circuit comprised of the following element: 81.59-ohm resistor, practical inductor with internal resistance of 0.99 ohm and reactance of 54.29 ohms; and a capacitor with reactance of 6.15 ohms. Compute for equivalent impedance angle in degrees. Note: Follow this reminder carefully. Compute to the nearest 4 decimal places. No Scientific notation. Do not round off in the middle of calculation. Use stored values. Write the numerical values only. No units in your final answer. Spaces are not allowed. Excessive number of decimals as compared to the required number of decimals may result to an incorrect answer.given: V1= 13.1V V2= 17V R1= 3.5 kilo-ohms R2= 6.4 kilo-ohms Solve the magnitude of the power (inmW) dissipated atR1termining the impedance and current of the system, the voltage across the resistor, and the voltage across the inductor. Could you please forward the response as a print of a handwritten answer? I got these two answers earlier. Which of them would be correct Impedance Z= (1000 + j 90.836) ohm Current I= 0.219∠-5.19° A Voltage across resistor VR = 219.097∠-5.19° V Voltage across inductor VL = 165.195∠84.81° V Impedance will be Z= 1004.052 ∠ 5.12° Current will be i = 0.221 ∠ -5.19° A. Volatge across resistor will be VR = 211.9∠ -5.19° V Volatge across inductor will be VL = 165.12 ∠ 84.81°V
- Calculate the difference A B if A=10236.87 and B=10453.13°. Show your solution graphically.9. The time constant of the circuit shown in the figure below is A. 25 sec. B. 0 sec. C. .025 sec. D. 0.25 sec. 50 다 2.5F 50The current waveform depicted in the given figure is characterized by a period of 4 s. i(i)(A) 4 hala... 3 4 5 6 3 2 1 7 8 t(s) What is the average value of the current over a single period? (You must provide an answer before moving to the next part.. The average value of the current over a single period is A.
- Source voltage from Channel 1 output of oscilloscope and VR(t) and VL(t) voltages from Channel 2 output are desired to be observed by applying TTL signal to a series RL circuit with inductance value RL=270 mH, series R resistance 2.6 K and inductance internal resistance Rin=430 Ω. . The settings for the oscilloscope are as follows. Channel 1 Volt/div adjustment 2V Channel 2 Volt/div adjustment 2V Time/div=0.1ms Since the time t=5τ is 4.4 square units, a) If VL(t) voltage is desired to be observed from Channel 2, draw the required circuit diagram with oscilloscope channel connections. b) Plot the VR(t) and VL(t) waveforms to scale according to the 4V TTL source condition, showing the 5τ and τ times. Compare the τ value you obtained with the τ value you will theoretically calculate. c) If the unit square number of the peak value on the vertical axis at the end of t=5τ of the VR(t) voltage seen in Channel 2 is 1.4, find the current passing through the series RL circuit. d) Explain…11 Consider the following circuit used to provide power for an inductive RL load. The input voltage is Vs=150V and the load has a 30 impedance value. The thyristor is working at a frequency fs = 2 kHz. The discharge current is to be limited to 50A and the required dv/dt is 50V/us. If the value of Cs is equal to 0.158µF, then the snubber losses are equal to: put of question T: T R Vs Select one: O a. 4.56W O b. 2.56W Oc. 5.45W Od. 3.56WApps > Google Play Maps Gmail YouTube Wolfram Alpha: Co.. Padlet: You are bea. C++ Tutorial for Be. Other bookmarks 国 If V = 23.8 V and C = 10.4 F, what is the charge (in C) of the capacitor between points 'a' and %3D b'? a 47 495.04 O 82.51 O 61.88 123-76 Time left 1:13:21 11:01 AM O Type here to search 25°C ENG 5/24/2021