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- H.W3 The single-phase full wave controlled rectifier has a R-L load with R= 0.552 and L= 6.5mH. The input voltage V, 220V at 50Hz. The delay angle is a-30, plot the input and output waveforms. Then determine: -The average thyristor current -The rms thyristor current -The output rms current -The average output current - The efficiency NOTE: Drive any formula that used in solutionVps = 10 V R = 0.1k02 ww -ovo R₁. Consider the Zener diode circuit shown in figure. The Zener diode voltage is Vz= 5.8V at Iz= 10mA and the Zener resistance is rz = 200. a) Find the output voltage for RL = 1k0 b) Find the change in the output voltage when the load resistance varies +ARL.in solution Yoad H.W2 The single-phase full wave controlled rectifier has a with R= 0.502 and L= 6.5mH. The input voltage V, 220V at 50Hz. The delay angle is a-60, determine: e) The average thyristor current f) The rms thyristor current g) The output rms current h) The average output current NOTE: Drive any formula that used in solution NG BRANCH
- Q4) Attempt to answer ONE branch only: a) Regarding ripple vector parameter, What are the differences between full wave and half wave rectifiers? b) From the circuit shown below which shows combining a positive clipper with a negative dipper Determine and draw the output voltage waveform? Note that Vp = 0.7V 1.h Vout 15 Vek 3 He V1in solution H.W3 The single-phase full waye ontrolled rectifier has a R-L load with R= 0.50 and L= 6.5mH. The input voltage V, 220V at 50Hz. The delay angle is a-30, plot the input and output waveforms. Then determine: -The average thyristor current -The rms thyristor current -The output rms current -The average output current - The efficiency - خطوات مل التقييم ا - رسم الدائرة . مرضيه let x=0 at 120Draw circuit diagram of a full-wave controlled rectifier and draw the wave shapes of output voltage and current for R-L load with sinusoidal input. You have to design a full-wave AC to DC converter circuit to have output average voltage of 40 and 50 W power to a 100 Ω resistive load from a source of 60 V. Is the design feasible? Why?
- A full-wave rectifier is connected to a single-phase source and supplying an inductive load. For the sake of simplicity, you can assume that the ripple at the output current is negligible. Input specifications: The source is single-phase utility grid at 220 V and 50 Hz. Output specifications: The load resistance is 10 Ohm. Sketch the schematic of the system. Sketch the waveforms of the output voltage and current and the source voltage and current. Determine all the performance parameters.My circuit is a Wall-Wall rectifier circuit. The transformer is taking in 120Vac from the wall and stepping it down to 22Vac. A set of 4 diodew and a capacitor then supply 22 VDC with a ripple voltage of 50mV and a ripple frequency of 120Hz. Which diode should be purchased and why?For the circuit in the figurea) Determine the time constant.b) Write the mathematical expression for IL, VL and VR, after the switch is closed.c) Determine IL, VL for one, three and five time constants.d) Draw the waveforms of IL, VL and VR.
- C4 R1 Vout For the precision rectifier circuit shown in Figure C4 what is the correct operation for the circuit if a Sine wave is 10k D1 R2 LM324 VEE applied to the input signal VIN? Vin D2 10k OUT V3 U1A Figure C4 A. When VIn is negative the circuit operates as unity voltage follower providing an in phase sine wave at VouT. When VIN is positive the circuit conducts but only to one diode drop, B. When VIn is negative the circuit operates as an inverting amplifier providing an inverted sine wave at VouT. When VIN is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, c. When VIn is negative the circuit operates as an inverting amplifier providing an in-phase sine wave at VOUT. When VIn is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, D. When VIn is negative the circuit operates as unity voltage follower providing an inverted sine wave at Vour. When VIn is positive the…1) Diode half-wave rectifier a) Consider the half-wave rectified sinewave voltage waveform at the right. What is the DC voltage and RMS voltage in terms of Vpeak? Remember: Vp peak ann T/2 1 Vay = DC = v(0)de avg T Sv(t)dt and V = Hve v² (t)dt T2. Connecting a capacitor at the output of a rectifier will serve as a a) Filter b) Charge storage device c) Bypass capacitor d) non of the above 3. In a half-wave rectifier , peak value of the output waveform is always a) Equal to the input wave b) Higher than the input wave c) Less than the input wave d) it depends upon diode 4. RMS value calculation involves which of the foolowing steps. a) Square root of all the values, taking mean, and then squaring b) Squaring all the values, taking means and then square root c) Squaring all the values, sqauare root them,and then taking means d) none of these 5. In full-wave rectifier, peak value of the output wave is always. a) Equal to the input wave b) Higher than the input wave c) Less than the input wave d) it depends upon the diodes