Q5/ An alternating voltage is given by voor) - 120 sin(314/-0.36) volts, find: A- The rms voltage; B- The periodic time; C- The time when the voltage is first a maximum.
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- Q2. For the Periodic Waveforms shown in the Figure, determine for each : Frequency, Average Value over half a cycle, RMS Value, Form Factor, and Peak Factor. 100 75 Voltage (V 50 25 20 40 Time (ms) 10 30 -100The peak distance of a sinusoidal waveform displayed on a C.R.O. screen is 11 cm and the 'volts/cm' switch is on 38 V/cm. The peak to peak voltage is given by O 3.45 V 295.53 V 836.00 V O 209.00 Vcan 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. NOTA
- The peak to peak distance of a sinusoidal waveform displayed on a C.R.O. screen is 9 cm and the 'volts/cm' switch is on 45 V/cm. The peak voltage is given by 405.00 V 202.50 V 5.00 V O 286.33 VAn rms voltage of 22.2 V with a frequencyof 1.00 kHz is applied to a 0.290-mH inductor. (a) What is the rmscurrent in this circuit? (b) By what factor does the current changeif the frequency of the voltage is doubled? (c) Calculate the current for a frequency of 2.00 kHzIn 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 10 V peak-to-peak. The rms value of this voltage is .What will be the rms value of this voltage if the frequency is reduced to the half value? 3.536 Vrms . 3.536 Vrms O 5 Vrms . 1.7677 Vrms O 3.536 Vrms. 1.7677 Vrms O 5 Vrms . 5 Vrms ...... O 10 Vrms . 10 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 AA voltage wave of 250 kHz has a Vmax-5 V and Vmin=-35 V. The duty cycle of this wave is 0.875. This voltage waveform has been applied across a 2.5 µH inductor for a very long time so that steady state has been achieved. Diagrams of the wave parameters and circuit are shown in this figure: L What is the average voltage of this wave? Average voltage = 35 V Voltage (arb, unit) 1.25 Minimum current: -3.5€ Amps 1 0.75 0.5 0.25 0 4.25 05 Have you integrated the wave over full cycle? 4.75 4 -1.25 Vmax 0.5 Vmin What is the average current flowing through the inductor? Average current = 3.56 A 15 Period/s On average, there is no power consumed by the inductor. Have you considered this? What are the maximum and minimum currents that flow through the inductor? Maximum current: 0.50 Amps D= 25 TH TH Have you used the relationship for current and voltage in an inductor? Have you considered the slope of the current wave?20 Find, how to relate the frequency and time period in a square waveform? a. not related b. square proportional c. directly proportional d. inversely proportional
- 1. A Sinusoidal current has an rms value of 5mA. Determine the following values. 1) Peak 2) Average 3) Peak to Peakiii) A sine waveform with following specifications. RMS voltage 161.4V Time for ten cycles 0.02s Sketch the waveform.When measured by an analog meter, a sinusoidal voltage shows a reading of 100 Vrms. What is the amplitude of the voltage source? A 140 V B) 141 V c) 142 V D 143 V