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- 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 140000for 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 vinFind the current through the load in polar form. R1 L1 R3 C1 100 150MH 250 30uF E E1 502 120 Vms 150 Vrms 60 Hz 60 Hz 30 45° 250m H
- Find the current in the resistor (R1) using the principle of superposition in the figure below. jón R. 62 -j82 V,=10260° V 1 JL=220 AjulSjrpjdLjcwDWXZoojA9XtU12Mi-pTxteUKH * In the circuit of figure below, v() is 7 V-4 ww 10 V () = 7F True False * In figure below, Vab=7 V in case (b) -5 5 V 5 V 5 V S V 3 V 3 V 3 V 3 V 1 V 1V 1V (a) (b) (c) (d) True FalsePlease show your detailed solution Number 14 The current in a series inductive circuit is 7.5 A at 25 Hz. The circuit takes 425 W, and the power factor is 0.47. The resistance of the circuit is A. 7.32 ohms C. 7.98 ohms B. 7.86 ohms D. 7.56 ohms
- sohw circuit diagram and step by step solution plsfor circuit shown in figure below find zj Vcc -20 V 6.8 k 220 k2 Ve Cc VB VoH Cc B- 180 Bo= 30 uS 56 k2 2.2 k2 CE P Type here to search4. A capacitor now is connected in parallel with R1 to reduce the ripple. Find the value of the capacitance needed to make the following ripple index less than 5% through PSpice simulation study. V -×100% -V o,min 0,max rpp RFP = -x100% = V dc dc D1 Dbreak R1 V1 1k VAMPL = 169.71 FREQ = 50 Vo
- Microelectrronics Theory Experts: Why is D1 open and D4 open, can you explain?-The Vollage V= 100210°V is 62 pplied across a capacilive 1oad . The 1-a kes ap paraal' Power Load of 500VA teading Power fracl-or. The reaclance (xc) of load 's---- 7Determine the nodal voltage V1 if l=4L25° A, and R=4 2, for the following figure: -j2 2 20/-90° V R Ref. Oa. 26.56 L 72.507 V Ob. 22.56 L 68.507 V Oc. 28.56 L 74.507 V Od. 24,56 L 70.507 V When the voltage source is compared with the current source, which of the following is true: Oa. The current source angle leads the voltage source angle by 135 Ob. The current source angle lags the voltage source angle by 115 Oc. The current source angle lags the voltage source angle by 135 Od. The current source angle leads the voltage source angle by 115 pe here to search DELL ww