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A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60° and output AC power of 2127 Watts, find the rectification efficiency
  • a)
    98.6 %
  • b)
    42 %
  • c)
    28 %
  • d)
    19 %
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A single-pulse transformer with secondary voltage of 230 V, 50 Hz, del...
Vo = (Vm/2π) x (1+cosα) = 77.64 V
Pdc = Vo2xR = 602.8 W
Rectification efficiency = Pdc/Pac = 28.32 %.
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Most Upvoted Answer
A single-pulse transformer with secondary voltage of 230 V, 50 Hz, del...
A half-wave controlled rectifier circuit, the output voltage is given by:

Vout = Vmax * sin(ωt)

Where Vmax is the maximum value of the input voltage, which is equal to the secondary voltage of the transformer, and ω is the angular frequency, which is equal to 2πf, where f is the frequency of the input voltage.

In this case, Vmax = 230 V and f = 50 Hz, so ω = 2π(50) = 100π rad/s.

To calculate the power delivered to the bulb, we need to find the average value of the output voltage. The average value of a sinusoidal waveform is given by:

Vavg = (2/π) * Vmax

So, Vavg = (2/π) * 230 = 146.45 V.

The power delivered to the bulb can be calculated using the formula:

P = V^2 / R

So, P = (146.45)^2 / 10 = 2130.61 W.

Therefore, the power delivered to the bulb is approximately 2130.61 W.
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A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60° and output AC power of 2127 Watts, find the rectification efficiencya)98.6 %b)42 %c)28 %d)19 %Correct answer is option 'C'. Can you explain this answer?
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