ELEX_4420_-_Lab_04_-_Single-Phase_FW_Rectifiers

A 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

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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 = 120

Topic: Full-Wave Rectification

92) For three- phase half-wave Controlled rectifier Shown in the figure: I- Derive. a formula for mean load Voltage. 2- Draw the wave forms of vo, Li,i, o and voltage across thyristor1 (r.) assuming highly inductive loud and d = 120 Vo

Full Wave Rectifier Circuit The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier.  Advantages of Full Wave Rectifier The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%. The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21. The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers. Question: What is the main disadvantage of a conventional full-wave…

a) Draw the 3-phase half-wave uncontrolled rectifier circuit for resistive load.b) Briefly explain the purpose of use.c) Draw the sinusoidal input voltages and the output voltage on the load.

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.

H.W:- For a single phase half wave uncontrolled rectifier if the input supply voltage is (v= 310 sin314t), at frequency (f= 50 Hz). Then determine the followings:- Vm. Vmean- i. Draw the rectifier circuit. un %3D ii. Draw the input and the output voltages waveforms. iii. Determine the mean and rms values of the output voltage. s Ec iv. If the load is (R=10 S), then determine the mean and rms values of the current. 2019

Facts Full Wave Rectifier Circuit The circuit of the full-wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full-wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier. Advantages of Full Wave Rectifier • The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%. • The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21. • The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers. Question: What is the main disadvantage of a conventional full-wave…

A single-phase full-wave rectifier is connected to an AC source with a peak voltage of 170 V and a frequency of 60 Hz. The load resistance is 502.

Draw the circuit diagram of a half-wave rectifier for producing a nearly steady dc voltage from an ac source. Draw two different full-wave circuits.

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3. For the half-wave rectifier given in the figure below, find the average values ofthe output voltage and current, and the ripple factor of the output voltage.

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 current

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.

3 Given the bridge-type full-wave rectifier circuit shown 220V 60Hz 220V:Es Es D1 D2 D3 D4 The load resistor is 220 and the transformer average current is 300mA. Determine: (a) Average value, RMS value, and frequency of load voltage VRL. (b) Average current and PIV of each diode. RL

A three phase full wave rectifier is shown below along with peak phase voltage Vm-169.7V. The load is purely resistive. The rectifier delivers Ipc = 100 A and the source frequency is 60 Hz. The DC output voltage is %3D VDc=280.7V and the output RMS voltage is equal to VRMS=280.93V. The efficiency, FF and RF are respectively equal to: Secondary D D D, R AD. Z D. D, Select one: O a. 99.83%, 100.08%, 4% O b. 99.83%, 55.08%, 4% O C. 99.83%, 100.08%, 16% O d. 87.83%, 100.08%, 4%

Which of the following statements about bridge tip rectifiers is false? A) If the 1-phase type has source inductance, the average value of the output voltage is lower than in the absence of the inductance. B) The average output voltage value of the 1-phase type is 0.9 times the peak voltage of the network. C) none D) There are only odd harmonics in the current drawn by the 1-phase type from the mains. E) The average value of the output voltage of the 3-phase type is larger as the 3-phase and 1-phase types are compared

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For the full wave rectifier circuit shown in Figure 1, calculate: 1) The maximum current through the load. 2) The average current through the load. 3) The average voltage across the load. 4) The rms value of the current. 5) The rectification efficiency. Vms = 280V Vin RL= 12ks2 Vrms = 280V Figure 1

An ohmic resistive load connected to the output of a single phase uncontrolled full-wave (with diode) AC / DC rectifier and the output voltage average value Va is obtained. Same resistance single phase full wave, controlled (with thyristor) AC / DC was connected to the output of the rectifier and the average value Vb of the output voltage was obtained. Va = 1.7 Vb Find the trigger angle of thyristors.

What condition should be followed to achieve discontinuous current mode operation in fully controlled rectifiers ? a. large L/R value b. large inductance value  c. rated inductance value  d. small L/R value