What is the necessary a.c. input power from the transformer secondary used in a half wave rectifier to deliver 500 W of d.c. power to the load?
A. 1232 W
B. 848 W
C. 616 W
D. 308 W
Answer: Option A
Solution (By Examveda Team)
Understanding Half-Wave Rectifiers:A half-wave rectifier only uses one-half of the input AC cycle to produce DC power. It essentially "chooses" the positive (or negative) part of the wave and discards the rest.
Power Loss:
Because it ignores half the input AC cycle, a significant portion of the input power is wasted as heat. This means the AC input power needed is much greater than the DC output power.
Calculating Input Power:
To find the AC input power, we need to consider this loss. For an ideal half-wave rectifier (no losses other than discarding half the wave), the relationship is approximately:
AC Input Power ≈ 2 x DC Output Power
Applying to the Problem:
The question states that the DC output power is 500W. Therefore:
AC Input Power ≈ 2 * 500W = 1000W
Why the options are different:
The options provided are slightly different from our simple calculation because real-world rectifiers have additional losses (like those in the transformer and diodes). However, our calculation gives us an approximation to help us pick the most likely answer. Considering the losses, the closest option is 1232 W.
Therefore, the best option is A.
This Question Belongs to Electronics And Communications Engineering >> Electronic Devices And Circuits
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Comments (5)
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40% efficiency of rectification does not mean that 60% of power is lost in the rectifier circuit. In fact, a crystal diode consumes little power due to its small internal resistance. The 100 W a.c. power is contained as 50 watts in positive half-cycles and 50 watts in negative half-cycles. The 50 watts in the negative half-cycles are not supplied at all. Only 50 watts in the positive half-cycles are converted into 40 watts
so input power *40% = output of halfwave rectifier
input power =500/(40/100)
=5000/4=1250
Given,
Dc load power = 500 W,
We need to find ac input power from the transformer.
As we know, the formula for the efficiency of half wave rectifier is as below,
efficiency = dc load power/ac input power.
We know that the efficiency of half-wave rectifier =40.6% or 0.406.
Therefore,
0.406=500/ac input power.
Ac input power = 500/0.406.
Ac input power = 1231.52.
How
40%= 500/Ac input power
Efficiency of half wave rectifire= DC load power/AC load power
40.5%