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The peak voltage in the output of a half-wavediode rectifier fed with a sinusoidal signalwithout filter is 10V. The d.c. component of theoutput voltage is [2004]
  • a)
    20/π V
  • b)
    10/√2 V
  • c)
    10/π V
  • d)
    10V
Correct answer is option 'C'. Can you explain this answer?
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Π V
b)5 V
c)10 V
d)0 V

Answer:
b) 5 V

Explanation:
The peak voltage of the half-wave rectified output is given by:

Vp = Vm

where Vm is the peak voltage of the sinusoidal input.

The average value (d.c. component) of the half-wave rectified output is given by:

Vdc = Vm/π

Therefore, in this case, Vdc = 10/π = 3.18 V

However, since the question asks for the peak-to-peak value of the d.c. component, we need to multiply this by 2:

Vdc(pp) = 2Vdc = 2(10/π) = 6.36 V

Therefore, the correct answer is b) 5 V, as this is the closest option to the calculated value of 6.36 V.
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The peak voltage in the output of a half-wavediode rectifier fed with a sinusoidal signalwithout filter is 10V. The d.c. component of theoutput voltage is [2004]a)20/πVb)10/√2 Vc)10/π Vd)10VCorrect answer is option 'C'. Can you explain this answer?
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