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network problems, please share your views!!

Printed From: One Stop GATE
Category: GATE Technical Discussions
Forum Name: GATE EE
Forum Discription: General Technical Discussions, Queries, doubts etc. for GATE in EE.
URL: http://forum.onestopgate.com/forum_posts.asp?TID=591
Printed Date: 08Feb2025 at 9:04pm


Topic: network problems, please share your views!!
Posted By: balu
Subject: network problems, please share your views!!
Date Posted: 22Feb2007 at 4:08pm
 GATE (91) PART B, PROBLEM-13

Soln:- The above problem has really puzzled me,
Equivalent ckt consists of a 50 source opposing a 10-60 source and series resistance of (6+4).
This gives i = (10-60 - 50 )/(6+4) = -j0.866 amps (anticlockwise sense)

Using voltage conjugate VI*,

Power 50 source………….( 50 )( -j0.866 )* (absorbed)
= j(4.33)

Power 10-60 source……….(10-60 )( -j0.866 )* (generated)
= 7.5 + j(4.33)

It indicates 7.5 W is dissipated in the (6+4).
Which gives, P6 = 4.5 W
P4 = 3.0 W

But the actual answer is different,

Power dissipated in the resistors = 0 (since real part of current is zero)

Power delivered by 50 source = 2.5W
Power delivered by 10-60 source = 10W





GATE (95), SECTION (B), Q-14,

Electromagnet problem…….

• How would u calculate  ?
max = 230/(4.44 f N)
rms = max/2

Or
230/Rg (as provided by Handa’s solution)

The answer does not come if the former is applied.




GATE (99) SEC-A, 1.9

What does “3db bandwidth of 5hz” mean in this question?






GATE (2000) SEC-B 4(b)

My answer: - B1/ (82)
Gate forum answer: - B1/8
Handa’s solution: - B1/ (22)



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