Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 16 (Q.No. 8)
8.
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?
Discussion:
7 comments Page 1 of 1.
Satya said:
4 years ago
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.
Therefore, the answer is 1232.
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.
Therefore, the answer is 1232.
(2)
Dharani said:
5 years ago
Please anyone explain it clearly.
Puneeth said:
5 years ago
The efficiency of hwr is 40.6%.
And the Ripple factor is 1.21.
And the Ripple factor is 1.21.
(1)
Arc said:
6 years ago
The efficiency is 40.6.
i.e 500/.406 = 1231.5.
i.e 500/.406 = 1231.5.
Renu said:
7 years ago
Here, hwr efficiency is 40.6.
Ajit said:
9 years ago
Given,
Dc load power = 500 W.
We know that efficiency = dc load power/ac input power.
For hwr efficiency = 0.402 or 40.2%
Therefore,
0.402 = 500/Pac.
Pac = 1250W
Dc load power = 500 W.
We know that efficiency = dc load power/ac input power.
For hwr efficiency = 0.402 or 40.2%
Therefore,
0.402 = 500/Pac.
Pac = 1250W
Unknown said:
1 decade ago
What is the solution for this?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers