Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A full wave rectifier uses load resistor of 1... Start Learning for Free
A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.
- a)368.98mW
- b)275.2mW
- c)145.76mW
- d)456.78mW
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A full wave rectifier uses load resistor of 1500Ω. Assume the di...
The AC power input PIN=IRMS2(RF+Rr).
IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA
So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.
IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA
So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.
Most Upvoted Answer
A full wave rectifier uses load resistor of 1500Ω. Assume the di...
10, and a peak input voltage of 10V. Calculate the peak output voltage and the average output voltage.
To calculate the peak output voltage, we first need to determine the voltage drop across the diodes. Assuming ideal diodes, this will be approximately 0.7V for each diode, so the total voltage drop across the two diodes will be 1.4V.
The peak output voltage can be calculated as follows:
Vout_peak = V_in_peak - 1.4V
V_in_peak = 10V (given)
Vout_peak = 10V - 1.4V = 8.6V
To calculate the average output voltage, we need to integrate the output voltage waveform over one cycle and divide by the period. The output voltage waveform for a full-wave rectifier with a load resistor can be approximated as a series of half-sine waves.
The formula for the average value of a half-sine wave is:
Vavg = (2 / π) * Vpk
where Vpk is the peak value of the half-sine wave.
For a full-wave rectifier, each half-sine wave has a peak value of Vout_peak / 2, so the average output voltage can be calculated as follows:
Vavg = (2 / π) * (Vout_peak / 2)
Vavg = (2 / π) * (8.6V / 2)
Vavg = 2.74V
Therefore, the peak output voltage is 8.6V and the average output voltage is 2.74V.
To calculate the peak output voltage, we first need to determine the voltage drop across the diodes. Assuming ideal diodes, this will be approximately 0.7V for each diode, so the total voltage drop across the two diodes will be 1.4V.
The peak output voltage can be calculated as follows:
Vout_peak = V_in_peak - 1.4V
V_in_peak = 10V (given)
Vout_peak = 10V - 1.4V = 8.6V
To calculate the average output voltage, we need to integrate the output voltage waveform over one cycle and divide by the period. The output voltage waveform for a full-wave rectifier with a load resistor can be approximated as a series of half-sine waves.
The formula for the average value of a half-sine wave is:
Vavg = (2 / π) * Vpk
where Vpk is the peak value of the half-sine wave.
For a full-wave rectifier, each half-sine wave has a peak value of Vout_peak / 2, so the average output voltage can be calculated as follows:
Vavg = (2 / π) * (Vout_peak / 2)
Vavg = (2 / π) * (8.6V / 2)
Vavg = 2.74V
Therefore, the peak output voltage is 8.6V and the average output voltage is 2.74V.
Free Test
FREE
| Start Free Test |
Community Answer
A full wave rectifier uses load resistor of 1500Ω. Assume the di...
Please sir i dont understanding this explaination
please help me in other way
please help me in other way
 | Explore Courses for Electrical Engineering (EE) exam |  |
Top Courses for Electrical Engineering (EE)View all
Top Courses for Electrical Engineering (EE)
Question Description
A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2026 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 full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer?.
A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2026 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 full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.a)368.98mWb)275.2mWc)145.76mWd)456.78mWCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
| Explore Courses for Electrical Engineering (EE) exam | |
Top Courses for Electrical Engineering (EE)
Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
