Problem 1 Apply the node-analysis by-inspection method to generate the node-voltage matrix for the circuit in the following figure and solve the problem. ΖΩΣ V 3Ω V2 14A Φιλ 3502
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- The characteristic equation of sixth order system is as below find the stability of system. + 2s° + 8s4 + 12s + 20s? + 16s + 16 = 0for the circuit shown below If v_in=5 sin2m10t then corner frequency will be ......Hz 1 UF 2 k Ohm 1 k Ohim 050 060 070 080 vinsohw circuit diagram and step by step solution pls
- K 1:08 docs.google.com 8 upload solutions here * Q1/ A. For the following Figune Find ITo tal, VRI, VR2 4kr siL Si3 1ov. 3 V R22K2 G1/B- For the following Figure VR3, VR2, IR2 , IR Find: R. lov a2/.Design Circint as the following pescripton: A- the positive Cycle of the vout signal is 3-2 and Negatine cycle is 4-3 (Note: we must used preetical Diodes and vin is tov) 13- the positive cycleg of vaut is s-3 and Negative cycle is (5.4) [Note:consider tovavd we must used Cupaetor with praetical Drode) vin = 1 Add file AYQ) By using suitable diagrum, describe the circuit laws below a) Kircho FF's la) KivchoFF'S Current lau Voltage lawA Thevenin de equivalent circuit always consists of an equivalent......... a. AC voltage source b. capacitance c. DC voltage source Cd resistance The superposition theorem is useful for the analysis of........... a. single-source circuits. b. only two-source circuits. d. no source circuits. multi-source circuits. If two currents are in the same direction at any instant of time in a given branch of circuit, the net current at that instant is the two currents ******** a. not sum of b. the product of c.) sum of d. the derivative of The Thevenin's equivalent circuit consists of a. Voltage source b. Ru c. Rth and Eth d all the mentioned answers D. In superposition theorem, when dealing with one source, the other sources are: a. If it's a current source, then it's short b. If it's a voltage source, then it's open circuit. circuit. If it's a current source, then it's open d. None of the mentioned answers. Cy circuit.
- Application: Construct the circuit in figure. R= 2,2KN, L = 1 mH, and C= 10 nF. L Necesssary Formulae: 1 Xc 2.π.f.C X.-2. π.f. L By using the formulea perform the calculations of Xc and XL then fill the table. f (Hz) Xc XL 50000 60000 70000 80000 90000 100000 110000 120000 130000 140000Vsupply1 10V (R1 22.2ko RC1 8200 C2 Vo C1 0.47µF HH B = 34 RL 4.7µF 24700 Vs R2 RE1 4700 1502 C3 #470µF Q1) (Dc analysis) For the circuit above find the following: 1. Ica- 2. VCEQ- 3. leQ- 4. VEQ- 5. IBQ- 6. VBQ- 7. Vca. Repeat the analysis above using: a) The approximate approach. b) The exact approach.The manimum value of the altenating wotaye and curent are 400 U and 20A respectvely s a cieritoanected to a svita supply . The ristantanerus values of the veltnge and current are 283V and 10 A respectvely of fime t=0, beth icreasing pesitively. 4) Write dan the gapressom for vwltaga and curend at pmet. b). Defermaned tha prver omsumed in the cranit. Take the velye and cunent to be ginusrida). fie P= VI end\ P = YI (PF)
- For the circuit given in fiqure 6, find V_ rms and ms on R3. R, R3 10 20 R2 RA 5 Vp Figure 6Solve V0 using nodal analysis by inspection. Show the equation for each cell Step-by-step, please. I need to understand the processFundamentals of Electrical Engineering 2020/2021 Dr. Yaseen H. Tahir Example: Find the equivalent conductance GAB for the circuit in Figure below and then find RAB- 2 mS ww 1mS A GAB - 36mS 3 mS 4mS: Solution: to CGAD= 8.493 MS R-1177.43o2 AB %3D