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 One Stop GATE Forum : GATE Previous Years Test Papers - Discuss Here : ME Papers |
 Topic: network problems, please share your views!! |
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manju
Senior Member 
Joined: 20Feb2007 Online Status: Offline Posts: 221 |
   Topic: network problems, please share your views!!Posted: 22Feb2007 at 10:56am |
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|  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) _________________ Thank you. Post Resume: Click here to Upload your Resume & Apply for Jobs |
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