JEE Advanced 2019 Cutoff for Admission in Electrical Engineering

JEE Advanced 2019 which will open gates of IITs for thousands of engineering aspirants is scheduled to be conducted on May 2019 after completion of the second phase of JEE Main 2019. This year approximately 2, 24, 000 students will appear for JEE Advanced 2019 across the country and the exam will be conducted by IIT Roorkee.

Electrical Engineering is one of the most sought after course and the cutoff of the top IITs are quite high.

Candidates need to qualify the JEE Main 2019 and should have a minimum of 75% in the 10+2 to fulfill JEE Advanced Eligibility. The exam will be conducted in one day through online mode and candidates will be selected for theTop IITs in India based on the ranks scored in the examination.

The shortlisted candidates will be offered various engineering branches according to their total marks scored and JEE Advanced Cutoffs.

What Determines IITs Cutoffs?

IITs cutoff varies based on a series of criteria, which also includes the popularity of the college, the number of seats available and the number of students applied for that IIT.  The cutoff varies from branch to branch as well. B. Tech in Computer Science and Engineering usually gets the highest rank preferences and the cutoff is higher followed by Electronics and Electrical Engineering.

Also Read: JEE Advanced 2019: Top IITs Opening and Closing Rank for Admission in Computer Science Engineering

Opening and Closing Ranks for Electrical Engineering:

The opening and closing rank of JEE Advanced 2019 will not be available before the release of the final result. However, there is a common trend in the ranks for the branches. Here is a list of expected cutoffs for Electrical Engineering branch in the IITs.

Also Read: JEE Advanced to clash with Lok Sabha Elections, changes will be available on jeeadv.ac.in