Us + 16 ΚΩ w 80 ΚΩ w 10 V -15 V Vo
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Given a range of values for vs and v0, how are these calculated for v0 and how is the range for vs made to avoid op-amp saturation if it is -2 V < vs < 3 V?
| vs(v) | 0.4 | 2.0 | 3.5 | -0.6 | -1.6 | -2.4 |
| vo(v) | -2 | -10 | -15 | 3 | 8 | 10 |
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- www H 0.47μF 14 V 68 ΚΩ 10 ΚΩ 1.2 ΚΩ Cwi Cwo 5.6 ΚΩ = 5 pF = 8 pF 0.47 μF HH B = 120 20 μF Cbc = 12 pF Cbe= 40 pF Cce = 8 pF V₂ 1. 3.3 ΚΩ a. Determine re- b. Find Avid Vo/Vi- = c. Calculate Z₁. d. Determine fife and fle e. Determine the low cutoff frequency. f. Determine fH, and fH.• V = 6.77 Volts C4 C₁ The values of the components are as follows: C₁ = 2.39 μFarads C₂ = 4.99 μFarads C3 = 7.42 µFarads C4 = 9.69 μFarads CA CB Fill in all the values into the table below: Element C (μF) V (V) C₁ C₂ CC 2.39 4.99 7.42 9.69 C₂ + HF V C3 {HD X 5.68 C4 X X Q (μC) U (μ)6V 520 Hz R₁ 100 2 www 2μF R2 470 C2 100 mH 1 μF
- 5. Find the indicated values using mesh analysis 30 R1 E1 5V 20 E2 10V 20° IL Polar Ic Polar IR PolarCalculate the power factor of the circuit below and draw the power triangle. The value of RNSID is 625.Inductive Circuits Fill in all the missing values. Refer to the following formulas: XL=2fLL=XL2ff=XL2L Inductance (H) Frequency (Hz) Inductive Reactance ( ) 1.2 60 0.085 213.628 1000 4712.389 0.65 600 3.6 678.584 25 411.459 0.5 60 0.85 6408.849 20 201.062 0.45 400 4.8 2412.743 1000 40.841