For the full-wave bridge rectifier circuit of Fig. 1, the ac source is 220 V rms at 60 Hz, R = 2 2, L = 10 mH, and Vdc = 80 V. Determine the power absorbed by the dc voltage source and the power absorbed by the load resistor.
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- For the below shown rectifier configuration, find the expression for the average value of voltage across the load resistor R, if the supply voltage Vs = Vm sinwt and firing angle = a. RLOAD Vm/T Cosa Vm/TT (1+cosa) O 2Vm/TT Cosa O 2Vm/TT (1+cosa) O O O OIn a phase controlled full wave rectifier, the source voltage is 50 VRMS sinusoidal, the source frequency is 60 Hz and the load is 20 ohms. Since the delay angle in the circuit is 20 °, which of the following is the value of the RMS load voltage? a) 99.54 V b) 49.77 V c) 70.38 V d) 35.19 V e) 23.46 V1. The full-wave controlled bridge rectifier shown has an ac input of 120 Vrms at 60 Hz and a 52 load resistor. The delay angle is 30°. Determine the average current in the load, the power absorbed by the load and the source voltamperes. V₁ I sin cor H www + Vo
- Q 2) A single phase half-wave uncontrolled rectifier with RL load is shown With R=1002, La 0. 1H w-377rad/s and V 100V, Determine: a) An expression for the load current i(t). b) The value of the current at t3D16 ms. c) The average load current (1C WLialx197=ア7.9 =121 V, = V, sinfar) 17.2 0.360Note: Answer All question 1: A) For a single-phase half-wave uncontrolled rectifier with (RL) load, find the rms value of the output voltage (V) and form factor (FF%) with supply voltage: v,(t) = 10 sin(wt). Draw the circuit diagram and sketch the voltages and current waveforms. The conduction angle B = 220°. %3DQ/3/A three phase controlled bridge rectifier is operated from a three phase Y- connected to a 250V, 50HZ supply. The rectifier is suppling a pure resistive load R=2002. Determine i) The required firing angle if average output voltage is 35% of the maximum output voltage ii)the average and r.m.s values of the load current ? iii) The average and r.m.s values of thyristor current
- Discussion : (1) رم طرة what are the advantăge and disadventage of full-wave bridge rectifier cutcuit- calculate ale practically result have been produced (2) Vpeak theortically of voltage output by using реа (3) üG Compare theortically and practically results. (4) What happens do on Vout, if we reversed the four diodes connects . sKetch the Vout in this case . 6. し 2.5 1.48 1.35Q1 a. Consider you have to choose b/w the Full-Wave Bridge Rectifier (FWBR) & Center-Tapped Full Wave Rectifier (CT-FWR) in one of your semester Lab projects. (1) Discuss the advantages of FWBR over CT-FWR which will justify, why to select the FWBR.A filtered 4 diode full wave rectifier circuit with the following specifications: The line load is 240 V rms with a frequency of 50 Hz, transformer ratio 199:20, with filter a 470 µF capacitor, and a 10KQ charge resistor. So: a. What is the transformer ratio if we want the ripple charge to be 0.3 V? b. If one of the diodes in the circuit is short, what are the possibilities?
- Three phase full wave bridge uncontrolled rectifier Va D1 D3 D5 Vb 1.5ko Vc D4 D6 D2 Fig.(1) Discussion: 1) Discuss clearly the operation of three phase full wave rectifier circuit. 2) Compare between three phase half wave rectifier and full wave rectifier. 3) Derive a formula for de and rms load voltage.Problem 1. Consider a half-wave rectifier circuit with a resistive load of 25 Ωand a 60 Hz ac source of 110V rms. Determine v0(t) and i0(t), the rms values of v0(t) and i0(t), the average power delivered to the load .4. The circuits parameters for an uncontrolled full-wave rectifier are as follows: Vin rms = 20V w= 314,159 rad/s Rload = 1KQ All diodes are Silisium and the equivalent circuit given below will be used. VFT = 0.7V, Ron = 0.01Q VET Ron + a) Find the value of DC voltage of the RL load resistance. b) Find the required rated PIV values for each diodes. c) Find the maximum conduction current of the diodes. d) Plot waveforms of the output current and voltage values.