GATE (91) PART B, PROBLEM-13
Soln:- The above problem has really puzzled me,
Equivalent ckt consists of a 50 source opposing a 10-60 source and series resistance of (6+4).
This gives i = (10-60 - 50 )/(6+4) = -j0.866 amps (anticlockwise sense)
Using voltage conjugate VI*,
Power 50 source………….( 50 )( -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 50 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/ (82)
Gate forum answer: - B1/8
Handa’s solution: - B1/ (22)
Which one is correct?
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