Detemine the value of w' in the figure such that the peak value of i(t) is 0.25 A. Use the identities z = VR2 +X², and – jx. =- if needed. jwc i(t) 10 A w- v(t) = 25 cos wt V F0.002 F None of the answers 6.67 10.2 5.02
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- Calibri (Body) -11 by AaBbCcDx AaBbCcD AaBbC AaBbCc AaB AaBbCc. mat Painter BIU 1Normal 1No Spaci. Heading 1 Heading 2 Title Subtitle Chang Styles d Font Paragraph Styles IMPEDANCE QUIZ # 2 1. What capacitance when connected in series with a 500Q resistor will limit the current drawn from a 48-mV 465-kHz source to 20µA? * a. 144pF b. 145pF c. 146pF d. 147pF MY ANSWER: c.146pF 2. What is the total impedance at 20kHz of a series circuit consisting of a 1.5mH inductance, a 100Q resistance, and a 0.08uF capacitance? * a. 1340 41.7 b. 1930 L-37.2" C. 920 290 d. 530 2-12.6°1. Use the Gram-Schmit procedure to obtain an orthogonal basis for the set of waveforms shown in Figure 1. 0.4T T. 0.4T 0.8T 0.6TFor the following parts, identify the name of each circuit, plot the output waveform, and find the DC output average voltage. Part (a) V 1 cycle V₁ = V sin cor Part (b) PE Part (c) Vm Vi ele eeeeeeeeeeeeee D D₂ D₁ D₂ R Va + D₂ D₁
- 5. Figure Q5 shows a parallel dipper circuit. Sketch the output waveform t, for one complete cycle. R Vi 10 V Si Si vo -10 V 7.3 V 5.3 V Figure Q5The peak distance of a sinusoidal waveform displayed on a C.R.O. screen is 6 cm and the 'volts/cm' switch is on 33 V/cm. The peak to peak voltage is given by O 396.00 V O 5.50 V O 139.99 V O 99.00 VShown in the figure below is an "RL" circuit drive by an AC power source. The AC power source has an RMS voltage of Vps (RMS) = 9.84 Volts and is running at a frequency of f = 8.585e+04 Hz. The resistor has a resistance of R = 2170 and the inductor has an inductance of L = 3.54e-03 Henries. Vps R ww Write the FORMULA for the total impedance of the circuit Ztot = Determine the numerical value of Ztot = 2890.5 Determine the numerical value of $z= = 41 Determine the current through the circuit: • I(PEAK) = 4.81E-3 • I(RMS) = 3.404E-3 Determine the voltage across the resistor: • VR(PEAK) = 7.387 • VR(RMS) = 5.22 ✔✔ Amps ✔Amps Write the FORMULA for the phase of the total impedance of the circuit z... = | tan-1 2701 R x Volts X Volts Determine the voltage across the inductor: • VL(PEAK) = 9.184 • VL(RMS) = 6.49 ✔ Volts Volts L R²+ WL- ✔ degrees 2 If a second circuit were connected in parallel with the inductor, this circuit would be considered as: O a low-pass filter O a capacitive switcher…
- can you please give me the comple solution for this question, thanks. A certain voltage function contains a constant term, a fundamental and third harmonic. The maximum value of the fundamental is 85 % of the constant term and the maximum value of the third harmonic is 55 % of the constant term. If the effective value of this function is 180.3V, find the magnitude of the two harmonics. A.IstHarmonic = 75 V , 3rd Harmonic = 120V B. Ist Harmonic = 80 V , 3rd Harmonic = 125V C. IstHarmonic = 120 , 3rd Harmonic = 75V D. Ist Harmonic = 125 V , 3rd Harmonic = 80V E. NOTAFrom step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?In an experiment, the function generator is adjusted to generate a square voltage at a certain frequency. This voltage is displayed on an oscilloscope and the reading is 16 V peak-to-peak. The rms value of this voltage is .What will be the rms value of this voltage if the frequency is doubled? Select one: O 16 Vrms 32 Vrms O 8 Vrms . 8 Vrms ..... O 5.657 Vrms 5.657 Vrms ....... O 8 Vrms . 16 Vrms O 16 Vrms 8 Vrms ......
- A pure sinusoidal current is being rectified. For the given maximum value of half wave rectified current is 50 A, then the rms value of full wave rectification will be 50 (a) A (b) 100 - A TC (c) 100 A (d)70.7 AQ.2) Determine all values of resistors and capacitor for the triangular waveform generator whose output alternates between +10Vpeak and -10V peak when the input is a 10-Hz square wave that alternates between +2V and -2V. The input resistance to the generator must be at least 15 kQ. you may assume that there is no de component or offset in the input.Determine the rms current in the figure for Vrms = 16 V and C = 74µF V f=10 kHz