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A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____ . (round off to nearest integer).
Correct answer is '800'. Can you explain this answer?
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A single-phase full-bridge diode rectifier feeds a resistive load of 5...
The output voltage of a single-phase full-bridge diode rectifier feeding a resistive load of 50 ohms can be determined using the following steps:
1. During the positive half-cycle of the input voltage, the diodes D1 and D2 will be forward biased, while D3 and D4 will be reverse biased. This allows the input voltage to pass through D1 and D2 and reach the load.
2. The peak output voltage can be determined using the peak input voltage, which is typically given in the problem. Let's assume the peak input voltage is Vpk.
3. The peak output voltage can be determined by subtracting the voltage drop across the forward biased diode from the peak input voltage. The voltage drop across a forward biased diode is typically around 0.7V.
So, the peak output voltage (Vout) can be calculated as:
Vout = Vpk - 0.7V
4. The average output voltage can be determined by taking the average of the rectified output waveform. Since the load is resistive, the average output voltage will be equal to the peak output voltage divided by pi (π).
Average output voltage (Vavg) = Vout / π
5. The RMS output voltage can be determined by taking the square root of the average of the squared output voltage waveform. Since the load is resistive, the RMS output voltage will be equal to the peak output voltage divided by sqrt(2).
RMS output voltage (Vrms) = Vout / sqrt(2)
6. The output current can be determined using Ohm's Law, where the output current (Iout) is equal to the RMS output voltage divided by the load resistance.
Output current (Iout) = Vrms / Load resistance
So, to summarize, the output voltage (Vout) can be calculated using the peak input voltage (Vpk) and the voltage drop across the diode, while the output current (Iout) can be determined using the RMS output voltage (Vrms) and the load resistance.
1. During the positive half-cycle of the input voltage, the diodes D1 and D2 will be forward biased, while D3 and D4 will be reverse biased. This allows the input voltage to pass through D1 and D2 and reach the load.
2. The peak output voltage can be determined using the peak input voltage, which is typically given in the problem. Let's assume the peak input voltage is Vpk.
3. The peak output voltage can be determined by subtracting the voltage drop across the forward biased diode from the peak input voltage. The voltage drop across a forward biased diode is typically around 0.7V.
So, the peak output voltage (Vout) can be calculated as:
Vout = Vpk - 0.7V
4. The average output voltage can be determined by taking the average of the rectified output waveform. Since the load is resistive, the average output voltage will be equal to the peak output voltage divided by pi (π).
Average output voltage (Vavg) = Vout / π
5. The RMS output voltage can be determined by taking the square root of the average of the squared output voltage waveform. Since the load is resistive, the RMS output voltage will be equal to the peak output voltage divided by sqrt(2).
RMS output voltage (Vrms) = Vout / sqrt(2)
6. The output current can be determined using Ohm's Law, where the output current (Iout) is equal to the RMS output voltage divided by the load resistance.
Output current (Iout) = Vrms / Load resistance
So, to summarize, the output voltage (Vout) can be calculated using the peak input voltage (Vpk) and the voltage drop across the diode, while the output current (Iout) can be determined using the RMS output voltage (Vrms) and the load resistance.
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A single-phase full-bridge diode rectifier feeds a resistive load of 5...
1 - φ full bridge diode rectifier,
Load, R = 50 Ω
1 - φ Active power supply
Vs = 200 Volts
P0 avg = V20rms/R
P0 avg = V20rms/R
V0r = Vs rms = 200 (for 1 - φ full bridge rectifier)
P0 avg = 2002/50 = 800 W
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A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____ . (round off to nearest integer).Correct answer is '800'. Can you explain this answer?
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A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____ . (round off to nearest integer).Correct answer is '800'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____ . (round off to nearest integer).Correct answer is '800'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____ . (round off to nearest integer).Correct answer is '800'. Can you explain this answer?.
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