in a full wave rectifier with input frequency 50 Hz the ripple in the output is mainly of the frequency (in Hz)
Correct answer: C. 100
- A. 25
- B. 50
- C. 100
- D. none
Explanation
In a full-wave rectifier circuit, the output waveform consists of both positive half-cycles of the input alternating current (AC) signal. As a result, the output frequency of the full-wave rectifier is double that of the input frequency.If the input frequency (f_in) is 50 Hz, the output frequency (f_out) of the full-wave rectifier will be:f_out = 2 * f_inFor the given input frequency of 50 Hz:f_out = 2 * 50 Hz = 100 HzHowever, the question asks about the frequency of the ripple in the output. The ripple frequency is equal to the output frequency of the full-wave rectifier, which is 100 Hz.So, the ripple in the output of a full-wave rectifier with an input frequency of 50 Hz is mainly of the frequency 100 Hz. The correct answer is (c) 100 Hz
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
Practise Electromagnetism
1,176 free Electromagnetism MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Electromagnetism questions
an inductor coil when consume no energy is called
the resistance of voltmeter must have very high resistance because
an electron of charge e coulomb passes through a potential difference of V volts. its energy in joules will be
The SI unit of magnetic flux is Weber, which is equal to:
The magnetic field due to current in solenoids can be increased by
Volt × second/ampere is equal to