The maximum efficiency of a half-wave rectifier is:

This question was previously asked in
MPMKVVCL Bhopal JE Electrical 24 August 2018 Official Paper
View all MPPGCL Junior Engineer Papers >
  1. 50.6%
  2. 40.6%
  3. 81.2%
  4. 40.2%

Answer (Detailed Solution Below)

Option 2 : 40.6%
Free
MPPGCL JE Electrical Fundamentals Mock Test
20 Qs. 20 Marks 24 Mins

Detailed Solution

Download Solution PDF

Concept

The efficiency of the rectifier is defined as the ratio of the DC output power to the AC output power.

For half-wave rectifier 

The DC output power is given by:

The AC output power is given by:

The efficiency is given by:

 

η = 40.6%

For full-wave rectifier 

The DC output power is given by:

\(P_{o(avg)}=​​V_{o(avg)}I_{o(avg)}\)

The AC output power is given by:

The efficiency is given by:

 

η = 81.2%

Latest MPPGCL Junior Engineer Updates

Last updated on Jul 18, 2025

-> MPPGCL Junior Engineer Notification 2025 has been released for various fields of post (Advt No. 3233).

-> MPPGCL has announced a total of 90 vacancies for Civil, Mechanical, Electrical, and Electronics Engineering (Junior Engineer).

->  Interested candidates can submit their online application form, from 23rd July to 21st August 2025. 

-> MPPGCL Junior Engineer result PDF has been released at the offiical website.

-> The MPPGCL Junior Engineer Exam Date has been announced.

-> The MPPGCL Junior Engineer Notification was released for 284 vacancies.

-> The selection process includes a Computer Based Test and Document Verification.

-> Candidates can check the MPPGCL JE Previous Year Papers which helps to understand the difficulty level of the exam.

More Diodes and Its Applications Questions

Hot Links: teen patti glory teen patti master gold download teen patti master list