A single phase half-wave rectifier is connected to a 300V RMS 50Hz AC supply. If the rectifier is used to supply a resistive load of 500 Ohms. Calculate the equivalent DC voltage developed across the load, the load current and power dissipated by the load
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- What is CEMF?For the half-wave rectifier, the source is a sinusoid of 120 V rms at a frequency of 60 Hz. The load resistor is 5 ohm. Determine (a) the average load current, (b) the average power absorbed by the load.a) Explain the operation of the controlled full-wave bridge rectifier shown in figure 3. b) The controlled full-wave bridge rectifier shown in figure 3 has a source 120 V rms at 60 Hz, R = 1092, L=20 mH, and a = 60°. Determine: i. An expression for the load current. ii. The average load current. iii. The power absorbed by the load. iv. Sketch the graph of the output waveform. Vs S₁ S4 Figure $3 $₂ Lo Vo
- A half-wave rectifier is to provide an averagevoltage of 50 V at its output.a. Draw a schematic diagram of the circuit.b. Sketch the output voltage waveshape.c. Determine the peak value of the output voltage. d. Sketch the input voltage waveshape.e. What is the rms voltage at the input?92) For three - phase half-wave Controlleed rectifier Shown in he figure: 1. Derive a fomu la for mean load Voltage. 2. praw the wave form of vo, i,,i, iy and voltage across thyristor 1 (r.) assuming highly inductive load and d = 120A single-phase half wave uncontrolled rectifier circuit is operating with a purely inductive load and a source voltage of 100 sin (277t) V. An ideal PMMC ammeter is connected in series with the inductive load reads 20 A. Calculate the distortion factor of the rectifier. Answer Choices: a. 0.866 b. 0.707 c. 0.577 d. 0.816
- A single phase – half wave controlled rectifier with freewheeling diode is supplying a load consistingseries connected a resistor and an inductance from a 70.7V (RMS), 50Hz sinusoidal AC source.The firing delay of the thyristor is 90° and the load values are R=10Ω, L=0.1 H. Define the loadcurrent expression and draw the load current by calculating for first two periods. And calculate theaverage values of the load voltage and current.A load is connected in the Full Bridge rectifier circuit (Four Diodes). Usedtransformer primary voltage value is 220 V (rms). The transformer winding ratio is 1:10.Using the 0.7 V model of the diode,Load Rating: 7 K Ohmsa) How many % of conduction does each diode branch stay on?b) Find the average values of the load voltage.c) Find the average values of the load current.d) Find the effective value of the load voltage.e) Find the effective value of the load current.In the circuit in the figure, an AC voltmeter will be made with full wave rectifier circuit structure. R = 50ohm and diodesResistance values in the direction of transmission are Rd = 100ohm. Inside of the DC ammeter to be used as indicatorthe resistance is very, very small. The AC mark to be measured is Vs = 100 Sinwt Volts.a- Draw the shape of the current passing through the DC Ammeter and calculate its maximum value.b- Find the average value of the current passing through the DC Ammeter.c- Analyze the voltage seen at the ends of diode D1 for both alternans.d- Find the Rms value of the voltage seen at the ends of the diode D1.
- 2. A single-phase single-diode (half wave) rectifier is connected to a 240 V, 50 Hz AC supply. Neglecting the diode volt drop, derive the instantaneous load current equation, determine the current waveform, mean load voltage, and the mean load current for a load of a. Pure resistor of 5 Ohm b. An inductance of 0.1 H in series with 10 Ohm resistor. (clue : B= 265° or 14.72 ms)Design a single-phase full wave bridge rectifier with input voltage of 100v and inductive load of 10ohm and 40mH a. Show the waveform relation between the I/P and O/P b. Calculate the average O/P voltage c. Calculate the rms currentFor the single phase half-wave rectifier, the source is a sinusoid of 120 V rms at a frequency of 50 Hz. The load resistor is 5ohm. Determine (a) the average load current, (b) the average power absorbed by the load and (c) the power factor of the circuit.