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For half-wave rectifier, the value of the direct current is given by x.Im , where Im is the peak to peak value of the input. Find the value of x?
[Write the answer up to two decimal point]
  • a)
    0.10
  • b)
    0.14
  • c)
    0.16
  • d)
    0.18
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
For half-wave rectifier, the value of the direct current is given by ...
Half-wave rectifier
A half-wave rectifier is a circuit that converts an alternating current (AC) signal into a direct current (DC) signal. It allows only half of the AC wave to pass through and blocks the other half.

Direct current value
The direct current value, denoted as x.Im, is the average value of the rectified output waveform. It represents the magnitude of the DC component of the signal.

Peak to peak value of the input
The peak to peak value of the input, denoted as Im, is the difference between the maximum and minimum amplitude levels of the AC input signal.

Calculation
To find the value of x, we need to calculate the average value of the rectified output waveform.

The average value of a waveform can be calculated by integrating the waveform over one complete cycle and then dividing it by the period of the waveform.

For a half-wave rectifier, the rectified output waveform is a half-cycle of the input waveform.

The average value can be given by the formula:
Average value = (1/T) * ∫(0 to T/2) Vm * sin(ωt) dt

Where:
- T is the period of the input waveform
- Vm is the peak value of the input waveform
- ω is the angular frequency of the input waveform

For a half-wave rectifier, the period T is equal to the time period of the input waveform.

The time period of an AC waveform is given by:
Time period (T) = 1/frequency

The frequency of the input waveform is the reciprocal of the time period.

Therefore, the time period T is equal to 1/frequency.

Since the input waveform is a sinusoidal waveform, the angular frequency ω can be calculated using the formula:
Angular frequency (ω) = 2π * frequency

Substituting the values of T and ω into the formula for the average value, we get:
Average value = (1/T) * ∫(0 to T/2) Vm * sin(ωt) dt
Average value = (1/(1/frequency)) * ∫(0 to 1/(2*frequency)) Vm * sin(2π * frequency * t) dt

Simplifying the expression and solving the integral, we get:
Average value = Vm / (2π * frequency) * [-cos(2π * frequency * t)] from 0 to 1/(2*frequency)
Average value = Vm / (2π * frequency) * (-cos(2π * frequency * (1/(2*frequency))) + cos(2π * frequency * 0))

Simplifying further, we get:
Average value = Vm / (2π * frequency) * (-cos(π) + cos(0))
Average value = Vm / (2π * frequency) * (-(-1) + 1)
Average value = Vm / (2π * frequency) * (2)

The peak to peak value of the input signal is given as Im.
The peak value of the input signal can be calculated as:
Peak value = Im / 2

Substituting the value of peak value into the expression for average value, we get:
Average value = (Im/2) / (2π * frequency) * (2
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Community Answer
For half-wave rectifier, the value of the direct current is given by ...
For Half wave rectifier we have, l0 is the peak value.
Ide = I0/π = 1/π(Im/2) = 1/2π.Im
Thus, x = 1/2π = 0.16
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For half-wave rectifier, the value of the direct current is given by x.Im , where Im is the peak to peak value of the input. Find the value of x?[Write the answer up to two decimal point]a)0.10b)0.14c)0.16d)0.18Correct answer is option 'C'. Can you explain this answer?
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